ra  MI 

\       Irving 

2MOEIAM 
Stringham 

GEOMETRY  FOR  BEGINNERS. 


BY 


G.  A.  HILL,  A.M. 


BOSTON: 

PUBLISHED   BY   GINN,    HEATH,   &  CO. 
1884. 


—  VV 


Entered  according  to  Act  of  Congress,  in  the  year  1880,  by 

GEORGE  A.  HILL, 
in  the  office  of  Jie  Librarian  of  Congress,  at  Washington. 


J.  S.  CUSHING  &  Co.,  PKINTERS,  101  PKAKL  STREET,  BOSTON. 


PREFACE. 


'  I  **HE  formal  method  of  teaching  Geometry,  as  we  find  it  in  Euclid 
JL  and  in  the  common  text-books,  whatever  may  be  its  merits,  is 
a  very  bad  method  for  beginners.  It  makes  the  study  of  the  subject 
unnecessarily  dry,  tedious,  and  difficult ;  and  it  ignores,  in  most  cases, 
the  great  law  of  mental  development,  that  clear  perceptions  and  intui- 
tions must  precede  the  intelligent  use  of  the  faculties  of  comparison 
and  reasoning.  The  usual  consequence  is  that  this  method  exercises 
the  memory  more  than  the  intelligence,  and  utterly  fails  to  impart  any 
habits  of  thought  that  are  useful  in  after-life. 

In  the  present  work,  a  method  is  employed  which  seems  to  the  author 
much  better  suited  to  promote  the  natural  growth  of  the  mental  powers. 
The  method  will  speak  for  itself  to  those  who  will  take  the  trouble  to 
examine  it.  A  marked  feature  of  it  consists  in  the  numerous  exercises 
which  are  to  be  worked  by  the  learner.  By  this  kind  of  work,  if  the 
exercises  are  well  chosen  and  faithfully  studied,  the  inventive  faculty  is 
called  into  play,  and  the  habit  of  seizing  quickly  the  true  relations  of 
things  is  acquired.  And  it  is  just  here,  —  in  its  power  to  form  this 
habit  of  mind,  —  that  the  value  of  the  study  of  Geometry,  as  a  prepar- 
ation for  the  varied  duties  and  labors  of  life,  can  hardly  be  over-esti- 
mated. Does  not  the  possession  or  want  of  this  habit  constitute,  in 
most  instances,  the  chief  intellectual  difference  between  a  man  who 
succeeds  in  the  world  and  a  man  who  does  not  ? 

In  Germany  the  true  worth  of  Geometry  as  an  educational  means  is 
better  recognized  than  elsewhere;  and  the  simpler  parts  of  the  subject, 


800542 


IV  PREFACE. 

treated  much  as  in  this  work,  have  been  introduced  with  the  happiest 
results  into  the  common  schools.  The  author,  while  residing  in  Ger- 
many in  1877  and  1878,  made  himself  familiar  with  their  methods  and 
text-books,  and  he  has  freely  used  the  knowledge  thus  acquired  in 
writing  the  present  work. 

As  regards  the  subjects  treated  in  the  last  foui1  chapters,  the  author 
was  not  satisfied  with  what  he  found,  even  in  the  best  German  text- 
books, and  he  has  planned  and  written  these  chapters  according  to  his 
own  ideas. 

To  make  the  study  both  more  interesting  and  more  useful,  much 
attention  has  been  given  to  the  practical  uses  of  Geometry ;  such,  for 
example,  as  measuring  inaccessible  distances,  and  computing  lengths, 
areas,  and  volumes. 

In  a  few  cases  the  method  of  Limits  has  been  used.  By  this  means 
rigor  of  proof  has  been  secured ;  and  when  the  method  is  presented  in 
its  simplest  form  (as  on  page  230) ,  the  author  believes  that  it  is  quite 
within  the  comprehension  of  the  average  boy  fifteen  years  old. 

If  answers  to  the  exercises  are  generally  desired  they  will  soon  be 
published. 

The  author  desires  to  express  his  obligations  to  Prof.  G.  A.  WENT- 
WORTH,  who  has  kindly  read  the  proof-sheets,  and  furnished  him  with 
many  valuable  suggestions. 

G.   A.    HILL. 
CAMBRIDGE,  June  i,  1880. 


CONTENTS. 


PAGE. 

CHAPTER   I.  — INTRODUCTION 1 

I.  Space  (§  i).  II.  The -Cube  (§§  2-5).  III.  The  Cylinder  (§  6). 
IV.  Bodies,  Surfaces,  Lines,  and  Points  (§§  7-15).  V.  Generation  of 
Space  Magnitudes  (§  16).  VI.  Straight  and  Curved  Lines  (§  17). 

VII.  Plane  and  Curved  Surfaces  (§  18).    VIII.  Cornered  and  Curved 
Bodies  ($  19).     IX.  Geometry  (§  20). 

CHAPTER   II.  — STRAIGHT  LINES 17 

I.  Direction  of  One  Line  (§$  21-25).  H.  Change  of  Direction.  The 
Circle  (§§  26-28).  III.  Directions  of  Two  Lines  (§§  29,  30).  IV.  Di- 
rections of  Three  Lines  (§  31).  V.  Length  of  a  Line  (§§  32-36).  VI.  Ra- 
tio of  Two  Lines  (§§  37,  38).  VII.  Units  of  Length  (§§  39-41). 

VIII.  Measuring  a  Line  (§§  42,  43). 

CHAPTER   III.  — ANGLES 51 

I.  Definition  of  an  Angle  (§  44).  II.  Magnitude  of  an  Angle  (§§  45, 
46).  III.  Magnitudes  of  Particular  Angles  (§§  47-49).  IV.  Measure 
of  an  Angle  (§§  50-52).  V.  Angles  made  by  Two  Lines  (§§  53,  54). 
VI.  Angles  made  by  Three  Lines  (§§  55-58). 

CHAPTER  IV.  — TRIANGLES 73 

I.  Sides  of  a  Triangle  ($$  59-63).  II.  Angles  of  a  Triangle  (§§  64-66). 
III.  Similarity,  Equivalence,  and  Equality  (§$  67,  68).  IV.  Equal  Tri- 
angles (§§  69-79).  V.  Some  Consequences  of  the  Equality  of  Triangles 
(§§  80-93).  VI.  Applications  (§§  94-98). 

CHAPTER  V.  —  QUADRILATERALS Ill 

I.  Sides  and  Angles  of  a  Quadrilateral  (§§  99,  100).  II.  Different  Kinds 
of  Quadrilaterals  (§§  101-105).  III.  Construction  of  Quadrilaterals 
(§§  106-111).  IV.  Subdivision  of  a  Line  (§§  112-114). 


VI  CONTENTS. 


PAGE. 

CHAPTER  VI.  — POLYGONS 123 

I.  Sides  and  Angles  of  a  Polygon  (§$  115-117).  II.  Regular  Polygons 
(§§  118-122).  III.  Symmetrical  Figures  (§  123). 

CHAPTER  VII.  — AREAS 136 

I.  Units  of  Area  (§§  124, 125).    II.  The  Areas  of  Polygons  (§§  126-135). 

III.  Practical  Exercises  and  Applications  (§§   136-141).     IV.  Theorem 
of  Pythagoras  (§$  142,  143).    V.  Transformation  of  Figures  (§§  144-150). 

VI.  Partition  of  Figures  (§§  151-155). 

CHAPTER  VIII.  — SIMILAR  FIGURES 171 

I.  Proportional  Lines  and  Figures  ($§  156-158).  II.  Similar  Triangles 
(§§  159-168).  III.  Problems  and  Applications  (§§  169-176).  IV.  Simi- 
lar Polygons  (§§  177-179). 

CHAPTER   IX.  — THE   CIRCLE 199 

I.  Sectors,  Angles  at  the  Centre,  Chords,  Segments  (§§  180-184).  II.  In- 
scribed Angles  (§§  185-188).  III.  Secants  and  Tangents  (§§  189-192). 

IV.  Two  Circles  (§$  193,  194).    V.  Inscribed  and  Circumscribed  Fig- 
ures   (§§    195-207).     VI.    Length   of  a   Circumference    (§§    208-211). 

VII.  Area  of  a  Circle  (§§  212-218). 

CHAPTER  X.  — THE  ELLIPSE 24O 

I.  The  Properties  of  the  Ellipse  (§§  219-221).  II.  Construction  of 
Ellipses  (§§  222-224). 

CHAPTER  XL  — PLANES 248 

I.  Straight  Lines  in  Space  (§§  225-227).  II.  A  Plane  (§§  228-230). 
III.  A  Plane  and  a  Straight  Line  (§§  231-235).  IV.  Two  Planes 
(§§  236-239).  V.  Three  Planes  (§  240).  VI.  Solid  Angles  (§  241). 

CHAPTER  XII.  — GEOMETRICAL  BODIES 263 

I.  Polyhedrons  (§§  242,  243).  II.  The  Prism,  Cylinder,  Pyramid,  and 
Cone  (§§  244-250).  III.  The  Sphere  (§§  251,  252). 

CHAPTER  XIII.  — SURFACES  AND  VOLUMES  OF  BODIES     .    279 
I.  Surfaces  of  Bodies  (§$  253-258).     II.   Volumes  of  Bodies  (§§  259- 
274).     III.  Exercises  and  Applications  (§§  275-279). 


GEOMETRY   FOR    BEGINNERS. 


CHAPTER  I. 
INTRODUCTION. 

CONTENTS.  — I.  Space  ($  i).  II.  The  Cube  (§$  2-5).  III.  The  Cylinder  (§  6). 
IV.  Bodies,  Surfaces,  Lines,  and  Points  (§§  7-15).  V.  Generation  of  Space 
Magnitudes  (§  16).  VI.  Straight  and  Curved  Lines  ($  17).  VII.  Plane  and 
Curved  Surfaces  (§  18).  VIII.  Cornered  and  Curved  Bodies  (§  19).  IX.  Geom- 
etry (§  20). 

L— Space. 

§  1.  We  cannot  define  space.  We  know,  however,  that  all 
things  exist  in  space.  I  lay  a  book  on  the  table  :  it  has  now  a 
definite  position  in  space  on  the  table.  The  table  is  in  the  space 
occupied  by  the  room ;  the  room  is  in  the  space  enclosed  by  the 
house ;  the  house  rests  upon  the  earth ;  and  the  earth  is  always 
moving  swiftly  through  space.  Thus  all  things  are  in  space  ;  but 
what  space  is,  where  it  begins,  or  where  it  ends,  these  are  questions 
which  no  man  can  answer. 

Space  contains  all  things,  and  extends  in  all  directions,  without 
beginning  and  without  end. 

II.— The  Cube.1 

§  2.  The  Cube  {fig.  /)  occupies  a  portion  of  space  which  is 
limited  on  all  sides.  A  limited  portion  of  space  is  called  a  BODY 
(see  §  7) .  Therefore  the  cube  is  a  body. 

1  A  model  of  the  cube  is  supposed  to  rest  on  the  table  or  other  support,  with 
one  face  turned  towards  the  eye  of  the  learner. 


2 


GEOMETRY    FOR    BEGINNERS. 


[is- 


The  cube  is  extended  in  space  in  three  chief  directions  :  from 
right 'to  left,  from  front  to  ^^/(%  from  <?^z;<? 
downwards.  We  express  this  fact  by  say- 
ing that  the  cube  has  three  DIMENSIONS, 
and  we  name  them  Length,  Breadth,  and 
Thickness  (Height  or  Depth).  Either  one 
of  the  dimensions  may  be  termed  length 
or  breadth  or  thickness,  but  the  terms 
height  and  depth  are  applied  only  to  the 
dimension  from  above  downwards. 


Fig.  i. 


The  three  dimensions  of  the  cube  are  equal  in  magnitude. 

Exercises.  —  1.  Here  are  several  cubes.  Have  they  all  the  same  shape? 
the  same  size  or  magnitude?  Are  they  all  composed  of  the  same  substance 
or  material ?  Why  are  they  called  bodies?  Why  are  they  called  cubes? 

2.  Mention  objects  which  are  cubical  in  shape,  or  resemble  a  cube. 
(Lumps  of  sugar,  pieces  of  soap,  tea-chests,  some  boxes,  etc.) 


§  3.  The  cube  is  limited  or  bounded  on  all  sides  by  a  SURFACE. 
When  we  handle  the  cube  it  is  its  surface  alone  which  we  touch. 
If  the  cube  is  not  transparent,  it  is  its  surface  alone  which  we 
are  able  to  see. 

The  entire  surface  consists  of  six  parts  or  partial  surfaces,  called 
its  Faces  :  the  upper,  lower,  front,  back,  right,  and  left  faces. 
Each  face  has  two  dimensions,  length  and  breadth  (height). 
Thus  the  upper  face  has  length  (extending  from  right  to  left)  and 
breadth  .(extending  from  front  to  back)  ;  the  front  face  has  length 
and  height,  etc. 

The  faces  of  the  cube  are  flat  or  plane  surfaces. 

Exercises.  —  1.  Name  the  faces  of  the  cube,  and  likewise  the  dimensions 
of  each  face. 

2.  Have  the  faces  of  the  cube  the  same  shape  or  form?  the  same  size? 
How  can  you  test  whether  they  have  the  same  size  or  not? 

Answer.  —  Lay  the  cube  on  a  piece  of  paper,  mark  around  its  base  with  a  pen- 
cil, then  apply  to  the  surface  thus  marked  out  the  other  faces  of  the  cube, 


§  4.]  CHAPTER    I. INTRODUCTION.  3 

3.  Give  examples  of  plane  surfaces  both  like  and  unlike  the  faces  of  the 
cube  in  form. 

4.  Do  the  faces  of  one  cube  have  the  same  shape  as  those  of  another  cube? 
the  same  size  ? 

§  4.  The  faces  of  the  cube  are  bounded  by  its  Edges.  These 
edges  are  LINES,  and  we  see,  (a]  that  each  face  is  bounded  by  four 
lines  or  edges,  (V)  that  each  edge  is  also  the  place  where  two  faces 
meet. 

The  edges  of  the  cube  have  only  one  dimension  :  length. 

The  edges  of  the  cube  are  straight  lines. 

Exercises.  —  1.    How  many  edges  has  the  cube  in  all? 

2.  If  the  cube  has  6  faces,  and  each  face  has  4  edges,  why  has  not  the 
cube  in  all  6  X  4=  24  edges? 

3.  Have  the  edges  of  the  cube  the  same  form?  the  same  size  (in  other 
words,  the  same  length")  ?     Test  whether  they  have  the  same  length. 

4.  Of  these  cubes,  which  has  the  longest  edges?     Which  the  shortest? 

§  5.  The  edges  of  the  cube  are  limited  by  its  Corners.  These 
corners  are  POINTS,  and  we  see,  (a)  that  each  edge  is  limited  by 
two  corners,  (b)  that  each  corner  is  the  place  where  three  edges 
meet. 

The  corners  of  the  cube  have  no  dimensions,  neither  length, 
breadth,  nor  thickness. 

How  many  corners  are  there  on  the  cube  ? 

Exercises. 

1.   This  -body  {Fig.  2,  I.)  is  a  three-sided  (or  triangular)  Prism  :  — 
(«)   Point  out  and  name  its  dimensions  (height,  length,  and  breadth). 

NOTE. — The  front  edge  gives  the  height  of  the  prism.  The  other  two  dimen- 
sions are,  (i)  the  distance  from  this  edge  to  the  opposite  face,  and  (2)  the  upper  or 
lower  edge  of  this  face ;  either  of  these  may  be  taken  as  the  length,  and  then  the 
other  will  be  the  breadth. 

(]})  How  many  faces,  edges,  corners,  are  there  ?  Which  are  surfaces, 
which  lines,  which  points?  (In  prisms  the  upper  and  lower  faces  are  called 
the  bases,  the  other  faces  the  lateral  faces.) 

r     Point  out  and  name  the  dimensions  of  each  face. 


GEOMETRY    FOR    BEGINNERS. 


[§  6. 


(X)  By  how  many  edges  (lines)  is  each  face  bounded? 
(e)   How  many  faces  meet  in  each  edge? 
(/)  How  many  edges  meet  at  each  corner? 

(g)  Have  the  faces  the  same  shape  (form)  ?  the  same  size  (magnitude)  ? 
(h)  Have  the  edges  the  same  form?  the  same  magnitude? 
(t)  Does  shape  or  size  or  material  determine  whether  a  body  is  a  three-sided 
Prism? 

2.  Examine  in  the  same  way  the  four-sided  (square)  Prism  (Fig.  2,  II.). 

3.  Examine  in  the  same  way  the  six-sided  (hexagonal)  Prism  (Fig.  2,  III.). 

4.  Examine  in  the  same  way  the  four-sided  (rectangular)  Pyramid  (Fig.  2, 


Fig.  2. 

III. —  The  Cylinder.^ 

§  6.  The  Cylinder  (Fig.  j)  is  also,  like  the  cube,  a  limited 
portion  of  space  :  in  other  words,  a  body. 
Its  three  dimensions  are  usually  called 
length,  breadth,  and  height.  The  height 
(the  straight  dotted  line  in  the  figure)  may 
have  any  value  :  so  may  the  length  and 
breadth,  but  they  must  be  always  equal  to 
each  other. 

The  cylinder  is  bounded  by  three  sur- 
faces :  two  plane  surfaces  called  its  Bases, 
and  a  curved  surface.     The  bases  of  the  cylinder  are  Circles. 


1  Several  models  of  right  cylinders  are  supposed  to  be  placed  on  the  teacher's 
desk  before  the  eyes  of  the  pupils. 


§7-] 


CHAPTER   I. INTRODUCTION. 


The  cylinder  has  only  two  edges  ;  these  form  the  boundaries  of 
the  bases,  and  are  called  Circumferences.  They  are  not  straight 
lines,  but  curved  lines. 

The  cylinder  has  no  corners. 

Exercises.  —  1.  Here  are  several  cylinders.  How  do  they  differ  from 
one  another?  In  what  do  they  agree? 

2.  Give  examples  of  objects  which  have  the  shape  of  cylinders  (a  pencil, 
stove-funnel,  etc.). 

3.  Examine,  as  regards  dimensions,  surface,  etc.,  the  Cone  (Fig.  4,  I.). 
4»    Examine  in  like  manner  the  Truncated1  Cone  (Fig.  4,  II.). 

5*    Examine  in  like  manner  the  Sphere  (Fig.  4,  III.). 


IV. — Bodies,  Surfaces,  Lines,  and  Points. 

§  7.  A  body,  in  common  language,  is  a  limited  portion  of 
space  filled  with  matter,  which  we  can  see,  touch,  handle,  etc.  : 
for  example,  a  pencil,  a  book,  the  table,  a  tree,  the  earth,  the  sun. 
But  in  Geometry  we  pay  no  regard  to  the  matter  of  which  a  body 
is  composed ;  we  study  simply  its  shape  and  its  size :  in  other 
words,  we  regard  it  simply  as  a  limited  portion  of  space. 

Definition.  — A  GEOMETRICAL  BODY,  or  SOLID,  is  a  limited  por-^ 
tion  of  space. 

I  Truncated  means  cut  off.     The  body  is  also  called  the  Frustum  of  a  cone. 


6  GEOMETRY    FOR    BEGINNERS.  [§  8. 

A  body  has  three  dimensions,  denoted  by  the  terms  length,  breadth, 
and  thickness  (height  and  depth). 

NOTE.  —  Either  dimension  may  be  called  length,  or  breadth,  or  thickness.  A 
dimension  estimated  from  below  upwards,  is  often  called  height,  as  the  height  of 
a  monument ;  a  dimension  estimated  from  above  downwards,  is  often  called  depth, 
as  the  depth  of  a  well.  Width  is  also  used  in  the  sense  of  breadth. 

§  8.  The  limits  or  boundaries  of  a  body  are  surfaces.  Surfaces 
exist  not  only  on  the  exterior  of  a  body,  they  may  also  be  thought 
of  as  existing  in  its  interior.  When  we  cut  an  apple  into  two  parts 
and  then  hold  the  parts  together,  the  common  boundary  between 
the  two  parts  is  a  surface.  But  without  actually  cutting  through  a 
body  we  may  imagine  a  surface  extending  in  any  direction  com- 
pletely through  the  body.  This  surface  would  divide  the  body 
into  two  parts  and  form  the  common  boundary  between  them. 
Every  body  is  divisible  into  parts,  and  the  common  boundary 
between  two  adjacent  parts  is  a  surface. 

Definition.  —  A  SURFACE  is  the  limit  of  a  body,  or  common 
boundary  between  two  adjacent  parts  of  a  body. 

A  surface  has  two  dimensions,  length  and  breadth  (height). 

§  9.  The  limits  of  a  surface  are  lines.  Lines  exist  not  only  as 
the  limits  or  boundaries  of  surfaces  :  we  can  also  think  of  them 
as  existing  anywhere  in  a  surface,  where  they  form  the  common 
boundaries  between  adjacent  parts  of  the  surface. 

Definition.  —  A  LINE  is  the  limit  of  a  surface,  or  the  common 
boundary  between  two  adjacent  parts  of  a  surface. 

A  line  has  one  dimension,  length. 

§  10.  The  limits  or  ends  of  a  line  are  points.  Points  exist  not 
only  at  the  ends  of  lines,  but  also  between  the  ends,  where  they 
form  the  common  boundaries  between  adjacent  parts  of  the  line. 

Definition. — A  POINT  is  the  limit  of  a  line,  or  common  bound- 
ary between  two  adjacent  parts  of  a  line. 

A  point  has  no  dimensions. 


§    II.]  CHAPTER   I. INTRODUCTION.  7 

§  11.  Bodies,  surfaces,  lines,  and  points  are  four  different  kinds 
of  things  relating  to  space  which  we  can  think  and  reason  about. 

Definition.  — Bodies,  surfaces,  lines,  and  points  are  called 
SPACE  CONCEPTIONS. 

The  first  three  :  bodies,  surfaces,  and  lines,  are  extended  in  space, 
and  hence  possess  shape  {form}  and  size  {magnitude}. 

Definition.  —  With  reference  to  form,  bodies,  surfaces,  and 
lines  are  called  SPACE  FIGURES  ;  with  reference  to  magnitude,  they 
are  called  SPACE  MAGNITUDES. 

The  point  has  no  extension,  therefore  neither  form  nor  magni- 
tude. 

§  12.  We  have  seen  that  two  faces  of  the  cube,  prism,  etc., 
meet  in  an  edge,  that  is  to  say,  in  a  line.  Whenever  two  surfaces 
meet  they  are  said  to  cut  or  intersect  each  other,  and  the  place  of 
meeting  is  a  line  called  their  Intersection.  This  line  lies  at  once 
in  both  surfaces,  or  is  common  to  both  surfaces. 

Again,  as  we  have  seen,  the  edges  of  the  cube,  etc.,  meet  at  the 
corners  ;  that  is  to  say,  in  points.  Whenever  two  lines  meet,  the 
place  of  meeting  is  a  point  common  to  both  lines,  and  is  called 
their  intersection. 

Definition.  —  The  place  where  two  surfaces  or  two  lines  meet 
is  called  their  INTERSECTION. 

The  intersection  of  two  surfaces  is  a  line. 

The  intersection  of  two  lines  is  a  point. 

§  13.  Divide  a  body  (for  example,  an  apple)  into  any  number 
of  parts ;  each  part  is  also  a  body,  smaller  of  course  than  the 
whole  body,  because  a  part  of  a  thing  is  less  than  the  whole. 

Grind  a  lump  of  salt  into  the  finest  particles  you  can,  so  fine 
that  you  cannot  distinguish  them  by  the  eye.  Under  a  micro- 
scope, each  particle  is  seen  to  be  what  it  really  is,  —  a  little  lump 
of  salt,  having  length,  breadth,  and  thickness  :  in  other  words,  a 


8 


GEOMETRY    FOR    BEGINNERS. 


[§ 


body.  In  short,  subdivide  a  body  in  thought  as  far  as  you  please, 
the  smallest  conceivable  part  is  also  a  body,  occupying  space,  and 
having  length,  breadth,  and  thickness.  We  could  never  arrive  at 
a  part  which  was  a  surface,  a  line,  or  a  point. 

In  like  manner,  every  part  of  a  surface,  however  small,  is  also  a 
surface  ;  and  every  part  of  a  line  is  also  a  line. 

T/ie  parts  of  a  body  are  bodies,  the  parts  of  a  surface  are  sur- 
faces, and  the  parts  of  a  line  are  lines. 

§  14.  A  surface,  therefore,  is  not  a  part  of  a  body.  Lay,  in 
thought,  any  number  of  surfaces  one  upon  the  other :  in  this  way 
you  would  never  obtain  a  body,  but  always  again  a  surface.  The 
common  boundary  which  separates  water 
from  oil  resting  on  it  (-Fig*5),  is  neither 
water  nor  oil  nor  any  other  substance  :  it 
is  not  a  body,  but  a  surface.  A  surface  is 
no  more  a  part  of  a  body  than  a  shadow 
is  a  part  of  the  wall  on  which  it  rests. 

A  line  is  no  part  either  of  a  surface  or 
of  a  body.  Lay  any  number  of  lines  side 
by  side,  the  result  is  neither  a  surface  nor 
a  body,  but  always  a  line.  The  line  which 
forms  the  common  boundary  between  a 


Fig.  5- 


red  surface  and  a  blue  surface  is  itself  neither  red  nor  blue,  nor 
has  it  any  other  color. 

Lastly,  a  point  is  no  part  of  a  line,  a  surface,  or  a  body.  In 
conversation  we  often  hear  the  word  "point"  used  in  the  sense  of 
a  surface  of  a  certain  extent.  Thus,  we  speak  of  the  "  point  of 
crossing"  of  two  or  more  roads.  Substitute,  in  thought,  for  these 
roads,  lines  absolutely  without  breadth  or  thickness  ;  then  their  in- 
tersection would  be  a  point,  as  this  word  is  understood  in  Geom- 
etry. 


§  i5-3 


CHAPTER   I. INTRODUCTION. 


Fig.  6. 


§  15.  Except  in  thought,  surfaces,  lines,  and  points  cannot 
exist  apart  from  bodies.  In  thought,  however,  we  can  detach 
them,  so  to  speak,  from  bodies,  and  study  their  forms  and  proper- 
ties apart  from  the  bodies  to  which  they  belong. 

To  aid  the  mind  in  reasoning  about  space  conceptions,  we  rep- 
resent them  to  the  eye  as  follows  :  — 

A  point  is  represented  by  a  fine  dot  made  with  a  pencil  on  paper 
{Fig.    6),  or   with  a  crayon  on  the 
blackboard  :  and  is  named  by  means 
of  a  letter  (A,  B,  C,  etc.). 

A  line  is  represented  by  the  fine 
continuous  trace  of  a  pencil  on  pa- 
per, or  of  a  crayon  on  the  black- 
board, and  named  either  by  two  let- 
ters, placed  one  at  each  of  its  ends 
(see  Fig.  7),  or  by  a  single  small 
letter  (a,  b,  c,  etc.). 

A  surface  is  represented  and  named  by  means  of  the  lines  which 
form  its  boundaries. 

A  body  is  represented  and  named  by  means  of  the  surfaces 
which  form  its  boundaries. 

Exercises.  —  1.    Give  examples  of  bodies. 

2.  Is  the  space  occupied  by  the  room  a  body?     Why? 

3.  Name  the  largest  body  you  can;   also  the  smallest. 

4.  Point  out  and  name  the  dimensions  of  the  room;   this  book;   this  pen- 
cil; this  piece  of  rubber. 

5.  Name  the  dimensions  of  a  board;   a  tower;  a  ditch  ;  a  well. 

6.  How  many  partial  surfaces  has  the  Voom?     Point  out  and  name  the  di- 
mensions of  each. 

7.  How  many  edges  has  the  room?    How  many  corners?     Of  what  are 
the  edges  the  intersections?   the  corners? 

8.  How  many  surfaces  has  this  piece  of  rubber?    Name  their  dimensions. 
Which  are  equal  in  magnitude? 

9.  A  glass  tumbler  has  an  exterior  (outer)  and  an  interior  (inner)  sur- 
face.    Give  other  examples  of  such  bodies. 


10  GEOMETRY    FOR    BEGINNERS.  [§    16. 

10.  How  many  edges  and  corners  has  a  pane  of  glass  ? 

11.  Cut  twice  through  this  body  (say,  an  apple)  so  as  to  divide  it  (if  pos- 
sible) into  two  parts;   three  parts;   four  parts;   five  parts. 

12.  Are  most  natural  objects  (for  example,  stones,  trees,  leaves,  flowers, 
animals)  simple  (like  the  cube,  prism,  etc.)  or  complex  in  shape  ? 

13.  When  we  use  such  an  expression  as  "a  shapeless  mass,"  do  we  mean 
that  the  mass  has  no  shape  at  all  ?     What  do  we  really  mean  ? 

14.  In  the  process  of  beating  gold,  leaves  are  obtained  not  over  g-cjuVfftf 
of  an  inch  thick.     Are  such  leaves  surfaces  or  bodies? 

15.  In  conversation,  we  often  hear  a  string,  a  telegraph  wire,  etc.,  spoken 
of  as  lines.     What  are  they,  strictly  speaking?     Why? 

16.  Is  the  trace  of  a  pencil-point  on  paper,  which  we  call  a  line,  really  a 
line?     What  is  it?     Why? 

17.  When  we  make  a  point  on  paper  with  the  end  of  a  pencil,  is  it  really 
a  point,  or  does  it  stand  for  or  represent  a  point?     What  is  it,  strictly  speak- 
ing?    Why? 

V.  —  Generation  of  Space  Magnitudes. 

§  16.  When  we  move  the  point  of  a  pencil  on  paper,  it  leaves 
behind  a  trace  which  we  call  a  line  :  reduce,  in  thought,  the  point 
of  the  pencil  to  a  point  as  understood  in  Geometry,  then  its  trace 
or  path  would  be  a  geometrical  line.  Again,  every  one  has  ob- 
served that  the  path  of  the  red-hot  end  of  an  iron  rod,  when 
moved  rapidly  in  the  dark,  presents  the  appearance  of  a  luminous 
line. 

The  path  of  a  moving  point  is  always  a  line  :  in  other  words, 
when  a  point  moves  in  space  it  describes  or  generates  a  line. 

When  a  line  moves  in  space,  it  usually  generates  a  surface. 
This  fact  may  be  illustrated  by  moving  a  crayon  over  the  surface 
of  the  blackboard,  with  its  longest  side  pressed  against  the  board. 

When  a  surface  moves  in  space,  it  usually  generates  a  solid. 
We  may  illustrate  this  truth  by  taking  a  thin  piece  of  board  (or  a 
slate)  to  represent  a  surface,  and  pressing  it  down  through  a  soft 
substance,  like  snow  :  the  hole  thus  made  is  the  path  of  the  sur- 
face, and  is  a  geometrical  body  or  solid. 


1 7-1 


CHAPTER'    I.  —  INTRODUCTION. 


11 


We  have  then,  in  all,  three  general  truths  or  laws  :  — 

1.  —  The  path  of  a  moving  point  is  a  line. 

II.  —  The  path  of  a  moving  line  is  a  surface, 

III.  —  The  path  of  a  moving  surface  is  a  solid, 

NOTE. —  Practically  speaking,  in  Law  I.  material  points  are  always  meant;  that 
is,  bodies  which  seem  to  be  points,  or  are  for  the  time  regarded  as  points.  The  stars, 
to  our  eyes,  are  material  points ;  in  reality,  they  are  bodies  of  enormous  size. 

Exercises.  —  1.  What  exception  is  there  to  Law  II.?  Illustrate  with  a 
pencil. 

2.  What  exception  is  there  to  Law  III.?     Illustrate  with  a  slate. 
3t    What  is  the  path  of  a  moving  solid? 


VI.  — Straight  and  Curved  Lines. 

§  17.  If  a  point  move  always  in  the  same  direction,  its  path 
is  a  straight  line  ;  if,  on  the  contrary,  the  direction  of  its  motion 
is  continually  changing,  its  path  is  a  curved  line  or  curve.  In 
Fig.  7,  A  B  is  a  straight  line,  CD  is  a  curve. 

Definitions.  —  I.  A  STRAIGHT  LINE  is  a  line  which  has  every- 
where the  same  direction. 

II.  A  CURVED  LINE,  or  CURVE,  is  a  line  whose  direction  is  con- 
tinually changing. 

A  line  like  E  F  (Fig.  7) ,  composed  A 
of  several   straight   lines,  is   called  a    / 
broken    line  :     and  a  line  like   G  H,  c 
composed    of    straight    and    curved 
lines,  is  called  a  composite  line. 

The  figures   on   carpets  and  wall- 
papers,  and  the   ornamental  designs  Q  ~  H 
employed  on  articles  of  furniture  and 


Fig.  7. 


in  architecture,  furnish  examples  of  broken  and  composite  lines 
in  endless  variety. 

The  word  line,  when  used  alone,  usually  means  a  straight  line. 


12 


GEOMETRY   FOR   BEGINNERS. 


[§ 


Exercises.  —  1.    On  the  blackboard  are  lines  a,  b,  c,  d,  etc.    What  kind  of 
a  line  is  a?     Why?     b  ?  c?  d?  etc.     Give  the  reason  in  each  case. 

2.  Draw  and  name  with  letters  two  lines  of  each  kind.     In  what  two  ways 
may  you  name  or  denote  a  line  ? 

3.  What  kind  of  lines  are  the  edges  of  the  cube?  the  prism?  the  pyra- 
mid? the  cylinder?    this  table?   this  ruler?   this  rubber?   the  room?   also  the 
dm  of  a  tumbler?  the  spokes  of  a  carriage-wheel?  its  tire?  a  telegraph  wire ? 
the  letters  M,  N,  O,  U,  V,  Z? 


Fig.  8. 

4.  What  sort  of  a  line  does  a  flash  of  lightning  often  resemble?     (Fig.  5.) 

5.  What  is  the  path  of  a  falling  apple?  a  ball  thrown  into  the  air?  the  end 
of  a  watch-hand?  the  end  of  a  pendulum?  a  drop  of  rain?  a  flying  bird?  a 
point  on  the  tire  of  a  carriage-wheel? 

6.  Give  other  examples  of  straight  lines  and  curves. 

7.  In  these  exercises  have  we  been  using  the  word  "line "in  its  stricl 
geometrical  sense?    Why? 

VIL  —  Plane  and  Curved  Surfaces. 

§  18.  The  faces  of  the  cube  are  flat  or  plane :  the  side  of  ^ 
cylinder  is  round  or  curved.  How  can  we  define  these  two  classes 
of  surfaces? 


§19.]  CHAPTER  I.  —  INTRODUCTION.  l3 

Definitions.  —  LA  PLANE  SURFACE,  or  PLANE,  is  a  surface  on 
which  straight  lines  can  be  drawn  in  all  directions. 

II.  A  CURVED  SURFACE  is  a  surface  on  which  straight  lines 
cannot  be  drawn  in  all  directions. 

A  straight  line  which  moves  in  space,  with  one  end  fixed  and 
the  other  end  sliding  along  another  straight  line,  generates  a  plane 
surface ;  but  if  the  second  line  is  a  curve,  the  surface  generated  is 
in  general  a  curved  surface. 

If,  for  example,  the  second  line  is  the  circumference  of  a  circle, 
the  surface  generated  will  be  that  of  the  cone  (Fig.  4),  except 
when  the  moving  line  is  in  the  plane  of  the  circle  :  in  this  case  the 
surface  generated  will  be  a  plane  surface. 

Test  of  a  Plane  Surface.  —  Apply  to  the  surface,  in  different 
directions,  the  straight  edge  of  a  ruler ;  if  the  surface  is  plane,  the 
edge  of  the  ruler  will  always  touch  the  surface  from  end  to  end. 

Exercises.  —  1.   On  a  tin  pail  point  out  a  plane  surface;  a  curved  surface. 

2.  What  kind  of  surfaces  are  the  faces  of  a  pyramid?  the  side  of  a  cone? 
its  base  ?  the  surface  of  the  sphere  ? 

3.  Are  the  walls  of  the  room  plane  or  curved?  the  surface  of  the  table? 
of  a  lamp-shade?  of  a  hat  ?  of  a  mirror?  of  a  wash-basin?  of  the  water  in 
the  basin?  of  a  flower?  of  the  ocean? 

4.  Give  other  examples  of  plane  and  curved  surfaces. 

5.  Can  you  draw  a  straight  line  on  the  side  of  a  cylinder?  on  the  side  of 
a  cone  ?  on  the  surface  of  a  sphere  ? 

6.  Test  whether  the  surface  of  your  desk  is  a  perfect  plane. 

VIII.  —  Cornered  and  Curved  Bodies. 

§  19.  The  cube  and  the  bodies  shown  in  Fig.  2  are  examples  of 
cornered  bodies.  The  cylinder  and  the  bodies  shown  in  Fig.  4 
are  instances  of  curved  bodies. 

Definitions.  — I.  A  CORNERED  BODY,  or  POLYHEDRON,  is  a  body 
bounded  entirely  by  plane  surfaces. 

II.  A  CURVED  BODY  is  a  body  bounded  partly  or  wholly  by 
curved  surfaces. 


14  GEOMETRY   FOR   BEGINNERS.  [§  2O. 

Exercises.  — 1.    Give  examples  («)  of  polyhedrons,  (<$)  of  curved  bodies. 
2.   To  which  class  does  the  earth  on  which  we  live  belong? 

IX.  —  Geometry. 

§  20.    The  Science  of  space  conceptions  is  called  GEOMETRY. 
The  aim  of  Geometry  is  threefold  :  — 

I. —  To  discover  the  various  properties  of  space  figures  arising 
from  their  having  different  forms . 

II.  —  To  show  how  to  draw  or  construct  space  figures. 

III.  —  To  show  how  space  magnitudes  may  be  measured. 
Geometry  is   usually  divided  into  Plane  Geometry  and  Solid 

Geometry. 

PLANE  GEOMETRY  treats  of  space  conceptions  confined  to  one  and 
the  same  plane. 

SOLID  GEOMETRY  treats  of  space  conceptions  not  confined  to  one 
plane  {for  example,  solids  and  curved  surfaces) . 

Chapters  II.  — X.  of  this  work  treat  of  Plane  Geometry;  the  re- 
maining chapters,  of  Solid  Geometry. 

NOTES. —  I.  The  word  Geometry  comes  from  two  Greek  words  meaning  land- 
measuring.  In  ancient  times,  when  the  science  was  in  its  infancy,  it  consisted  en- 
tirely of  a  few  practical  rules  employed  in  measuring  land,  and  used  especially  by  the 
Egyptians  in  settling  disputes  about  boundary  lines  destroyed  by  the  annual  over- 
flow of  the  river  Nile.  Hence  the  name  Geometry  (or  land-measuring)  was  given 
to  the  science  by  the  Greeks. 

2.  No  science  excels  Geometry  in  the  number  and  importance  of  its  practical 
applications.  This  fact  is  strikingly  seen  in  the  power  which  it  gives  us  of  measur- 
ing space  magnitudes.  If  one  asks :  How  much  more  will  grow  on  this  field  than 
on  that  one  ?  how  much  must  this  ship  be  loaded  to  sink  ten  feet  into  water  ?  how  to 
find  the  right  way  across  the  pathless  ocean  ?  how  maps  of  the  land  and  sea  are 
made  ?  how  the  distance  of  the  moon  can  be  found  ?  at  precisely  what  instant  an 
eclipse  of  the  sun  will  take  place  ?  when  a  comet  will  again  be  seen  ?  how  a  cannon 
must  be  pointed  in  order  that  the  ball  may  hit  a  certain  mark?  —  these  and  similar 
questions,  only  a  man  skilled  in  Geometry  can  answer. 


CHAPTER   I.  —  INTRODUCTION.  15 

REVIEW   OF   CHAPTER   I. 
QUESTIONS. 

1.  How  many  dimensions  has  the  cube,  and  what  are  they  called? 

2.  How  many  faces,  edges,  and  corners  has  the  cube? 

3.  How  many  faces  meet  in  an  edge?  how  many  edges  in  a  corner? 

4.  Are  the  faces  equal  or  unequal?  the  edges? 

5.  Answer  the  preceding  questions  for  each  of  the  bodies  in  Fig.  2. 

6.  What  account  can  you  give  of  the  surfaces  and  edges  of  the  bodies  in 
Figs.  3  and  4? 

7.  Name  a  body  which  has  no  corners;   also  one  which  has  neither  edges 
nor  corners. 

8.  Define   a  solid,  a  surface,  a  line,  a  point;    how  many  dimensions  has 
each,  and  what  are  they  called? 

9.  Define  the  intersection  of  surfaces  or  of  lines.     What  is  the  intersection 
of  a  surface?  of  a  line? 

10.  What  are  space  conceptions  ?  space  magnitudes  ?  space  figures  ? 

11.  What  are  the  parts  of  a  body?  the  parts  of  a  surface?  the  parts  of  a  line? 

12.  Illustrate  the  facts,  that  a  surface  is  no  part  of  a  body,  and  that  a  line  is 
no  part  of  a  surface. 

13.  Illustrate  what  is  meant  by  a  point  in  Geometry. 

14.  How  are  space  conceptions  represented  and  named? 

15.  Illustrate  how  space  magnitudes  are  generated  by  motion.     What  are 
the  general  laws?  the  exceptions  to  them? 

16.  Define  the  four  kinds  of  lines,  and  give  examples. 

17.  Define  plane  and  curved  surfaces.     How  is  a  plane  surface  tested? 

18.  Show  how  a  line  must  move  to  generate  a  plane  surface;   a  curved  sur- 
face. 

19.  Define  polyhedrons  and  curved  bodies,  and  give  examples. 

20.  What  is  Geometry  ?  its  chief  aims  ?  its  chief  divisions  ? 

EXERCISES. 

1.  Give  examples  of  bodies  shaped  like  a  prism,  a  pyramid,  a  cone,  a  sphere. 

2.  Name  the  dimensions  of  a  telegraph  wire,  a  monument,  a  river,  a  well. 

3.  Name  the  dimensions  of  a  cylinder,  (a]  when  standing  on  its  base,  (£) 
when  resting  on  its  side. 

4.  How  can  you  cut  a  prism  in  two  so  that  each  part  will  also  be  a  prism? 


16  GEOMETRY   FOR   BEGINNERS. 

5.  How  can  you  place  two  cylinders  having  equal  bases  together  so  as  to 
form  another  cylinder? 

6.  Can  you  place  two  cubes  together  so  as  to  form  a  new  cube? 

7.  Here  are  several  equal  cubes:  put  them  together  so  as  to  form  anew 
cube.     How  many  cubes  does  it  take?     How  much  larger  is  the  new 
cube  than  one  of  its  parts?  its  face  than  the  face  of  one  of  its  parts? 
its  edge  than  the  edge  of  one  of  its  parts?  • 

8.  Mention  bodies  of  regular  shape,  but  differing  in  shape,  from  those  thus 
far  examined. 

9.  What  kind  of  space  magnitude  is  the  human  skin? 

10.  Give  an  instance  where  more  than  two  lines  intersect  in  the  same  point. 

11.  Can  more  than  two  surfaces  intersect  in  the  same  line?    Can  you  give  an 
illustration? 

12.  When  will  a  large  body  appear  very  small?    Give  examples. 

13.  Draw  (if  possible)  a  line  with  no  ends  ;   with  one  end ;   with  two  ends ; 
with  more  than  two  ends. 

14.  If  you  slide  a  straight  line  along  the  two  rails  of  a  railroad,  what  kind 
of  a  surface  would  be  generated? 

15.  If  you   move   a   straight  line   around  the  tires  of  two  carriage-wheels 
on  the  same  axle  (always  keeping  it  in  contact  with  both  tires),  what 
kind  of  a  surface  would  be  generated? 


§   2!.] 


CHAPTER   II. STRAIGHT    LINES. 


17 


CHAPTER  II. 
STRAIGHT   LINES. 

CONTENTS.  — I.  Direction  of  One  Line  (§§  21-25).  H.  Change  of  Direction. 
The  Circle  (§§  26-28).  III.  Directions  of  Two  Lines  (§§  29,  30).  IV.  Direc- 
tions of  Three  Lines  (§  31).  V.  Length  of  a  Line  (§§  32-36).  VI.  Ratio  of 
Two  Lines  ($$  37,  38).  VII.  Units  of  Length  (§$  39-41).  VIII.  Measuring  a 
Line  (§§42,43). 

L  — Direction  of  One  Line. 

§  21.   A  straight  line,  A  B  (Fig.  9) ,  may  be  regarded  as  hav- 
ing either  the  direction  from  A  towards     A B 

B,  or  the  opposite  direction  from  B  to-          ^ 

wards  A;  in  the  first  case,  we  should 
call  it  the  line  AJ3,  in  the  second  case, 


Fig.  9. 


the  line  BA.  Sometimes  (as,  for  instance,  on  a  guide-post)  an 
arrow-head  or  a  hand  is  used  to  point  out  which  direction  is  in- 
tended ;  but,  in  general,  we  may  assign  to  the  line  either  direc- 
tion. 

§  22.  If  we  know  only  the  direction  of  a  straight  line,  we  can- 
not determine  its  position  or  locate  it ; 
for,  through  different  points,  A,  B,  C, 
D,  (Fig.  id),  different  straight  lines 
may  be  drawn,  all  having  the  same 
direction. 

Likewise,  if  we  know  only  that  a 
straight  line  passes  through  a  known  or 
given  point,  A,  the  line  is  not  known  ;  Fi&' I0' 

for,  through  a  point  any  number  of  straight  lines  may  be  drawn. 


18  GEOMETRY   FOR   BEGINNERS.  [§   23- 

But  through  the  point  A,  in  a  given  direction,  as  that  of  the 
arrow,  only  one  straight  line  can  be  drawn.  And,  also,  through 
any  two  points,  as  A  and  M,  but  one  straight  line  can  be  drawn. 

These  two  truths  may  be  stated  as  follows,  in  a  general  form, 
without  reference  to  Fig.  10,  or  to  any  particular  case  :  — 

1.  —  Through  a  given  point  in  a  given  direction  only  one  straight 
line  can  be  drawn. 

II.  —  Throttgh  two  given  points  only  one  straight  line  can  be 
drawn. 

Hence,  if  one  point  and  the  direction  of  a  line  are  given,  or  if 
two  points  of  the  line  are  given,  the  line  is  determined  in  position. 

§  23.  For  drawing  straight  lines  on  paper,  or  on  the  blackboard, 
a  RULER  (Fig.  7/)1  with  a  straight  edge  is  employed. 

In  order  to  draw  a  line  on  paper  through  two  given  points,  lay 
the  ruler  on  the  paper  so  that  its  edge 
just  touches  the  two  points  ;  then  draw 
the  line  with  a  hard,  well-sharpened 
pencil,  holding  the  pencil  nearly  vertical,  and  always  in  contact 
with  the  edge  of  the  ruler. 

When  we  draw  a  line  without  the  aid  of  any  instrument  to 
guide  the  hand,  we  are  said  to  draw  it  FREE-HAND. 

Exercises.  —  1.    Draw  four  lines  through  a  point,  and  name  them  with  letters. 

2,  Make  two  points,  and  then  join  them  by  a  straight  line. 

NOTE.  —  If  A  and  B  are  two  points,  the  expression  join  AB  \slo  be  understood 
as  meaning  draw  a  straight  line  from  A  to  B. 

3*  Make  three  points  not  in  the  same  line;  then  join  them  by  straight  lines. 
How  many  straight  lines  in  all  can  be  drawn? 

4.  How  many  straight  lines  in  this  way  can  be  drawn  between  four  points? 
five  points?  six  points? 

5.  Ttrzw,  free-hand,  a  straight  line;   then  correct  or  rectify  it  with  the  aid 
of  a  ruler.     Repeat  the  exercise  several  times. 

1  The  divided  rule  (see  $  42),  which  the  pupil  requires  for  other  purposes, 
will  also  serve  for  drawing  straight  lines. 


§  24.]  CHAPTER   II.  —  STRAIGHT   LINES.  19 

6.  How  can  you  test  whether  the  edge  of  a  ruler  is  straight? 

Ans.  —  Draw  a  line  with  the  ruler  through  any  two  points;  -then  turn  the  ruler 
end  for  end,  and,  along  the  same  edge,  draw  a  line  again  through  the  same  two 
points.  If  the  edge  is  straight,  the  two  lines  will  coincide  and  form  a  single  line. 
(See  §  22,  II.).' 

7.  Test  the  edge  of  your  ruler  or  divided  rule. 

8.  Test  the  edge  of  the  ruler  at  the  blackboard. 

9.  A  farmer  has  a  four-sided  field,  and  wishes  to  run  a  straight  fence  be- 
tween two  opposite  corners.     Explain  how,  with  a  person  to  assist  him,  he 
can  find  points  on  the  proposed  line  of  the  fence  between  the  corners. 

Suggestion. —  If  the  farmer  stands  a  little  way  behind  one  corner,  and  places 
his  eye  so  that  the  line  of  sight  passes  through  both  corners,  this  line  will  coincide 
with  the  proposed  line  of  the  fence. 

§  24.    Among  the   directions  which   a  straight  line  may  have, 
there  are  two  of  more  practical  importance  than  the 
others. 

Fasten  a  weight  to  one  end  of  a  cord,  and  then  hold 
the  other  end  at  rest  in  the  hand  :  this  forms  what  is 
called  a  plumb  line.  When  the  cord  is  at  rest  its  di- 
rection is  said  to  be  VERTICAL.  On  the  other  hand, 
a  pencil,  stick,  or  similar  object,  when  floating  on 
the  surface  of  still  water,  is  said  to  have  a  HORI-  ^ 
ZONTAL  direction.  Fis- I2- 

Definitions.  —  LA  VERTICAL  LINE  is  a  line  which  has  the 
direction  of  a  plumb  line. 

II.  A  HORIZONTAL  LINE  is  a  line  which. has  the  direction  of  any 
line  in  the  surface  of  still  water. 

III.  Lines  neither  vertical  nor  horizontal  are  called  INCLINED 
LINES. 

Plane  surfaces,  also,  are  either  vertical,  horizontal,  or  inclined. 
Every  plane  in  which  a  vertical  line  can  be  drawn,  is  a  vertical 
plane ;  the  surface  of  still  water,  or  any  plane  similarly  placed 
with  reference  to  the  earth's  surface,  is  a  horizontal  plane ;  planes 
neither  vertical  nor  horizontal  are  inclined  planes. 

The  sides  of  the  room,  for  example,  are  vertical  planes ;  the 


20  GEOMETRY   FOR   BEGINNERS.  [§  24. 

floor  and  ceiling  are  horizontal  planes ;  the  roof  of  the  house  is 
an  inclined  plane. 

NOTES.  —  i.  The  surface  of  water,  if  of  considerable  extent  (the  ocean,  for  in- 
stance), is  sensibly  a  curved  surface,  because  the  earth  is  round.  This  curvature 
makes  itself  evident  on  the  seashore  when  we  watch  a  ship  sailing  out  of  sight  on 
the  horizon  :  first  the  hull  sinks  out  of  sight ;  then  the  sails  and  lower  parts  of  the 
masts ;  last  of  all,  the  tops  of  the  masts.  But  on  a  pond  three  or  four  miles  in 
length,  the  curvature  is  so  small  that  it  cannot  be  observed ;  hence,  for  all  common 
purposes,  the  surface  of  the  pond  is  regarded  as  a  plane. 

2.  On  paper  draw  vertical  lines  towards  or  from  you ;  horizontal  lines  from  right 
to  left,  or  left  to  right. 

Exercises.  —  1.  Of  the  lines  on  the  blackboard,  which  are  vertical?  which 
horizontal?  and  which  inclined? 

2.  Draw  a  line  on  paper;   then  hold  the  paper  so  that  the  line  shall  be 
vertical.     Is  now  the  plane  of  the  paper  also  vertical? 

3.  Hold  a  pencil  vertical;   horizontal;   inclined. 

4.  Hold  a  book  vertical;   horizontal;   inclined. 

5.  Draw  three  lines  of  each  kind. 

6.  Draw  a  vertical  line,  mark  five  points  on  it,  and  through  these  points 
draw  horizontal  lines. 

7.  On  the  cube,  point  out  vertical  edges  and  faces;   also  horizontal  edges 
and  faces. 

8.  What  directions  have  the  edges  of  the  prism?  those  of  the  pyramid? 
those  of  the  cylinder? 

9.  What  direction  has  a  ladder  when  leaning  against  a  wall?  when  lying 
on  the  ground?  when  suspended  from  one  end  .by  a  rope?     What  direction 
have  the  rounds  of  the  ladder  in  these  three  cases? 

10.  What  direction  has  the  mast  of  a  ship?  the  path  of  a  falling  apple? 
the  beam  of  a  balance  at  rest?    the  surface  of  a  lake?    the  side  of  a  house? 
its  roof? 

11.  At  what  time  of  the  year  are  the  sun's  rays  most  nearly  vertical  ? 
When  are  the  sun's  rays  horizontal? 

12.  Give  other  examples  of  the  vertical  and  horizontal  lines  and  planes. 

13.  When   is    the   minute-hand    of  a  clock  vertical?    when  horizontal? 
Answer  the  same  questions  for  the  hour-hand. 

14.  When  two  vertical  planes  (for  example,  the  two  sides  of  a  room)  in- 
tersect each  other,  what  kind  of  a  line  is  their  intersection? 

15.  When  a  vertical  plane  intersects  a  horizontal  plane  what  kind  of  a 
line  is  their  intersection?     Give  an  example. 

16.  Through  a  point  how  many  vertical  lines  can  be  drawn?  how  many 
horizontal  lines?  how  many  inclined  lines?  how  many  planes  of  each  kind? 


§   25.]  CHAPTER   II.  —  STRAIGHT   LINES.  21 

17.  In  a  vertical  plane  how  many  vertical  lines  can  be  drawn?  how  many 
horizontal  lines? 

18.  In  a  horizontal  plane  how  many  vertical  lines  can  be  drawn?  how 
many  horizontal  lines? 

§  25.  Applications.  —  I.  In  order  to  test  whether  a  line  or  a 
plane  (as  a  wooden  beam,  the  wall  of  a  house,  etc.)  is  ver- 
tical, a  PLUMB-RULE  (Fig.  13)  is  often  used.  It  consists  of 
a  piece  of  wood  having  two  straight  edges  everywhere  equally 
distant  from  each  other.  Near  one  end,  exactly  half  way 
between  the  edges,  a  plumb-line  is  attached.  Near  the 
other  end,  also  exactly  half  way  between  the  edges,  a  mark 
is  made.  If  we  place  one  edge  of  the  plumb-rule  against  a 
surface  which  is  vertical,  the  line,  when  at  rest,  will  cover 
the  mark ;  if  the  surface  is  not  vertical,  the  line  will  pass  on 
one  side  of  the  mark.  It  is  well  to  apply  the  test  twice, 
using  first  one  then  the  other  edge  of  the  plumb-rule ;  in 
both  cases,  the  line  ought  to  cover  the  mark. 

II.  To  test  whether  a  line  or  a  plane  surface  is  horizontal  (or, 
as  workmen  say,  level),  instruments  called  LEVELS  are  used.  The 
simplest  kind  of  level  is  the  Plumb-line  Level  (Fig.  14) .  It  con- 
sists of  three  pieces  of  wood  put  together  in  the  shape  of  the  let- 
ter A,  with  a  plumb-line  suspended  from  the  top.  In  order  to 
adjust  the  instrument  for  use,  a  mark  must  be  made  on  the  cross- 
piece,  where  the  plumb-line  passes,  when  the  feet  of  the  level  are 


Fig.  14.  Fig.  15. 

at  the  same  height,  or  horizontal.     This  is  done  as  follows  :  Place 
the  feet  of  the  level  on  any  two  points,  and  mark  on  the  cross-bar 


22 


GEOMETRY   FOR   BEGINNERS. 


[§25. 


the  place  of  the  plumb-line ;  then  turn  the  instrument  end  for 
end,  rest  it  on  the  same  points,  and  again  mark  the  place  of  the 
plumb-line.  The  point  midway  between  the  two  marks  is  the 
right  one. 

Fig.  is  shows  the  instrument  in  use  to  test  whether  a  wooden 
beam  is  horizontal.  As  we  see  in  the  figure,  the  plumb-line  passes 
a  little  to  the  left  of  the  mark  on  the  cross-bar.  Which  end  of  the 
beam  is  the  highest? 

In  testing  whether  a  surface  (as  that  of  a  table)  is  horizontal, 
we  must  apply  the  level  in  at  least  two  different  directions,  because 
a  line  in  one  direction  on  the  surface  might  be  horizontal  while 
lines  in  other  directions  were  inclined.  But  if  two  lines  on  the 
surface  in  different  directions  are  horizontal,  then  the  surface  must 
be  horizontal. 

NOTES.  —  i.  A  common  Spirit-level  is  shown  in  Fig.  16.     It  is  more  accurate 

than  the  plumb-line  level,  but  (when  properly 
made)  more  expensive.  The  essential  part  of 
the  spirit-level  is  a  glass  tube,  slightly  curved, 
and  filled  with  alcohol,  except  the  space  oc- 
cupied by  a  bubble  of  air,  which  always  seeks 
Fig.  16.  the  highest  part  of  the  tube. 

2.  Surveyors  and  engineers,  in  establishing  horizontal  lines  on  the  ground,  or 
prolonging  them,  use  levels  mounted  either  on 
a  staff  which  is  driven  into  the  ground,  or  on 
a  three-legged  stand  (tripod)  which  rests  on  the 
ground.  One  may  make  a  simple  (but  not  very 
accurate)  plumb-line  level  for  use  on  the  ground, 
as  follows  :  Fasten  a  common  carpenter's  square 
{Fig.  77)  in  a  slit  in  the  top  of  a  staff  by  means 
of  a  screw,  and  then  tie  a  plumb-line  at  the 
angle  so  that  it  may  hang  beside  one  arm. 
When  it  has  been  made  to  do  so  by  turning 
the  square,  the  other  arm  will  be  horizontal. 
Fig.  77. 

Exercises.  —  1.  Choose  some  object  in  the  room,  and  test  whether  it  is 
vertical. 

2.  Choose  some  object  in  the  room,  and  test  whether  it  is  horizontal. 

3.  Draw  on  the  blackboard  (with  the  aid  of  a  plumb-rule)  a  vertical  line. 

4.  Draw  on  the  blackboard  (with  the  aid  of  the  plumb-line  level)  a  hori- 
zontal line. 


26.] 


CHAPTER  II.  —  STRAIGHT   LINES* 


II.  —  Change  of  Direction.    The  Circle. 

§  26.  The  direction  of  a  straight  line  is  susceptible  of  change. 
The  hands  of  a  clock,  for  instance,  may  be  regarded  as  two  lines 
which  are  constantly  changing  in  direction.  The  minute-hand,  in 
the  course  of  an  hour,  points  in  all  possible  directions  contained 
in  the  plane  of  the  face  of  the  clock. 

When  a  straight  line,  A  B  (Fig.  18) ,  moves  without  leaving  the 
point  A,  until  it  has  another  direction, 
as  A  C,  it  is  said  to  turn  or  ROTATE 
about  the  point  A.  If  the  motion 
continues,  the  rotating  line  will  at 
length  return  to  its  first  position,  A  B; 
it  is  then  said  to  have  made  one  REV- 
OLUTION. 


Fig.  18. 


§  27.  If,  during  the  motion  about  A,  the  length  of  the  line  re- 
mains unchanged,  its  other  end,  B,  will 
describe  a  curved  line  which  returns  into 
itself,  and  has  the  property  that  all  its 
points  are  at  the  same  distance  from  A. 
This  curved  line  (Fig.  19)  is  called  a 
CIRCUMFERENCE  ;  the  point  A  is  called 
its  CENTRE.  Of  all  curved  lines,  the 
circumference  is  the  simplest  and  the 
most  important. 

Definitions.  —  I.  A  ciirved  line  which  is  everywhere  equally 
distant  from  a  point  is  called  a  CIRCUMFERENCE. 

II.  The  point  is  called  the  CENTRE. 

III.  A  part  of  a  circumference  is  called  an  ARC  (for  example, 
B  C  or  CD,  Fig.  19). 


GEOMETRY   FOR   BEGINNERS. 


[§  28. 


IV.  A  straight  line  joining  the  centre  to  the  circumference  is 
called  a  RADIUS  1  (for  example,  AB  or  A  C,  Fig.  19). 

V.  The  plane  surface  bounded  by  a  circumference  is  called  a 
CIRCLE. 

NOTE.  —  The  word  circle  is  also  used  in  certain  expressions  in  the  sense  of  cir- 
cumference ;  thus,  in  the  expressions,"^  draw  a  circle"  "arcs  of  circles"  strictly 
speaking,  not  the  circle  but  the  circumference  or  boundary  of  the  circle  is  meant. 

From  definitions  I.  and  IV.  it  follows  that  — 
All  radii  of  the  same  circle  are  equal  in  length. 

Exercises.  —  1.  Name  objects  on  which  circles  and  circumferences  present 
themselves  (ends  of  a  cylinder,  rim  of  a  tea-cup,  a  hoop,  etc.). 

2.  What  does  the  tire  of  a  carriage-wheel  represent?    the  spokes?    the 
axle  ?  the  part  of  the  tire  between  two  spokes  ? 

3.  In  Fig.  18,  as  A  B  rotates  about  A,  what  kind  of  paths  do  points  be- 
tween A  and  B  describe? 

§  28.  Circles  (or,  speaking  more  exactly,  circumferences  and 
arcs)  are  drawn  on  paper,  etc.,  by  the  aid  of  an  instrument  called 
the.  COMPASSES  or  DIVIDERS  2  (Fig.  20). 

The  compasses  (for  use  on  paper)  consist  of  two  metal  legs 
connected  together  by  a  pivot  about  which 
they  can  turn.  The  legs  have  fine,  hard 
steel  points.  The  lower  part  of  one  leg 
can  be  removed  (by  turning  a  screw  on 
the  side)  and  its  place  supplied  by  a  pencil- 
leg,  or  a  leg  provided  with  a  steel  pen. 
In  Fig.  20  the  pencil-leg  is  shown. 

To  draw  a  circle  with  a  given  radius,  say 
two  inches,  fit  the  pencil-leg  to  the  com- 
Fig.  20.  passes,  open  the  legs  till  the  distance  apart 

of  their  points,  or  opening,  is  just  two  inches,  and  then  proceed 
as  shown  in  Fig.  20. 


1  Plural,  radii. 

2  So  called  from  its  use  in  dividing  lines  and  angles  into  equal  parts. 


§28.] 


CHAPTER  II. — STRAIGHT   LINES. 


25 


On  the  ground,  circles  are  often  described  by  the  aid  of  a  cord. 
When  a  gardener,  for  instance,  wishes  to  trace  a  circumference,  he 
proceeds  as  illustrated  in  Fig.  21. 


Fig.  21. 


Exercises.  —  1.    Describe  how  to  draw  a  circle  of  given  radius  on  paper. 

2.  Describe  how  a  gardener  traces  a  circle  on  the  ground. 

3.  Draw  (on  paper)  a  circle  with  a  radius  equal  to  the  breadth  of  your 
ruler  or  divided  rule. 

4.  Draw  a  circle  with  the  large  compasses  on  the  blackboard. 

5.  Can  you  draw  a  circle  on  the  blackboard  with  a  radius  greater  than  the 
greatest  opening  of  the  compasses? 

6»    Draw  several  arcs  of  circles. 

7.  Draw,  free-hand,  several  circles.     Which  is  drawn  the  best? 

8.  Draw  three  circles  all  having  the  same  centre,  or  concentric  circles,  as 
they  are  called.     The  waves  caused  by  throwing  a  stone  into  water  are  exam- 
ples of  concentric  circles  ;   what  is  their  common  centre  ? 

9.  Draw  (a)  any  circle;    (£)  a  circle  Math  a  point  A  as  centre,  and  any 
radius;    (<:)  a  circle  with  a  radius  of  one  foot,  and  in  any  position  (that  is,  hav- 
ing any  point  as  centre) ;    (of)  a  circle  with  the  point  A  as  centre,  and  a  ra- 
dius of  one  foot. 

10.  What  two  things  completely  determine  the  position  and  magnitude  of 
a  circle? 


26  GEOMETRY   FOR   BEGINNERS. 


III. — Directions  of  Two  Lines. 

§  29.   Two  straight  lines  in  the  same  plane  must  have  either 
(a)  the  same  or  (^)  different  directions. 
Case  I.  — When  two  different   lines,  as  AB  and   CD  (Fig. 

A —B    22},  have  the  same  direction,  they  are 

called    PARALLEL   lines.      Bearing    in 

mind  what  was  said  in  §  21,  we  may 

also  regard  one  of  them  as  having  one 

G  ~~ H    direction,  and  the  other   the  opposite 

Fig.  22. 

direction.  It  is  also  evident  that  sev- 
eral lines,  as  A  B,  CD,  F  F,  G  H,  may  be  parallel  to  one  another. 

Definition.  —  PARALLEL  LINES  are  lines  which  have  the  same 
direction. 

That  a  line  AB  is  parallel  to  a  line  CD  is  sometimes  ex- 
pressed, for  the  sake  of  brevity,  by  a  sign  ;  thus  :  A  B  ||  CD. 

The  two  rails  of  a  railroad  furnish  a  familiar  illustration  of  par- 
allel lines  ;  substitute  for  the  rails  geometrical  lines,  and  these  would 
be  parallel  lines  as  understood  in  Geometry.  The  opposite  edges 
of  a  table  are  another  example  of  parallel  lines. 


Fig.  23. 


Every  one  has  observed^that  the  rails  of  a  railroad  course  along 
without  either  approaching  or  separating.     We  may  suppose  them 


§  29-] 


CHAPTER    II. STRAIGHT    LINES. 


27 


Fig.  24  gives 


to  be  prolonged  as  far  as  we  please,  without  meeting.  The  same 
is  true  of  the  edges  of  the  table  and  of  all  lines  parallel  to  each 
other.  In  fact,  if  two  parallel  lines  could  meet  when  prolonged, 
they  would  then  be  two  straight  lines  passing  through  the  same 
point  (the  point  of  meeting),  yet  not  coinciding;  but  this  is  im- 
possible, for  all  straight  lines  drawn  through  the  same  point  in  the 
same  direction  must  coincide  and  form  only  one  line  (see  §  22, 1.) . 
Therefore,  — 

Parallel  lines  can  never  meet,  however  far  prolonged. 

But  two  lines  which  would  not  meet  if  prolonged  are  not  neces- 
sarily parallel.  For  instance  :  an  edge  of  the  table  and  one  of  the 
legs  on  the  opposite  side  would  never  meet  if  prolonged,  yet  they 
are  not  parallel.  The  lines,  in  order  to  be  parallel,  must  lie  in  the 
same  plane. 

Two  curves,  also,  may  be  parallel  to  each  other, 
an  example.  The  curves  would  not  meet  if  pro- 
longed, and,  at  corresponding  points,  as  A  and  B, 
or  C  and  D,  they  have  the  same  direction.  When  a 
railroad  makes  a  curve,  the  rails  do  not  cease  to 
be  parallel.  Concentric  circles  supply  another  ex- 
ample of  parallel  curves.  (See  Fig.  190,  p.  212.) 

Two  planes,  or  a  plane  and  a  straight  line,  are 
parallel,  if  they  would  not  meet  however  far  they 
were  extended.  The  floor  and  ceiling  of  the 
room  is  an  instance  of  parallel  planes ;  the  floor 
and  an  edge  of  the  ceiling  is  an  example  of  a  plane  and  straight 
line  parallel  to  each  other. 

Case  II.  —  Far  more  common  is  the  case  where  the  two  lines 
have  different  directions.    Two 
such  lines,  as  A  B  and   C D  Q-'~-~^. 
(Fig.  25} ,  either  actually  meet, 
or,  if  prolonged  in  one  or  the 
other  direction,  will  approach 


24> 


28  GEOMETRY    FOR    BEGINNERS.  [§   3°- 

and  at  last  meet  in  a  point,  (9,  called  their  intersection  (§  12). 
Two  edges  of  a  table  which  meet  at  a  corner  are  an  example 
of  lines  having  different  directions ;  the  corner  is  their  intersec- 
tion. 

Exercises.  —  1.  Which  lines  on  the  blackboard  are  parallel?  which  not 
parallel?  Prolong  the  latter  until  they  meet. 

2.  Can  two  lines,  not  parallel,  intersect  in  more  than  one  point? 

3.  How  do  you  read  the  expression  AB  ||  CD? 

4.  Hold  two  pencils  (#)  parallel;    (b~)  so  that  they  would  intersect;    (V) 
so  that  they  are  neither  parallel  nor  would  they  intersect. 

5.  Hold  two  books  (a)  parallel ;    (U}  not  parallel. 

6.  Hold  a  pencil  and  a  book  (#)  parallel ;    (b*}  not  parallel. 

7.  Point  out  on  the  cube  (a)  parallel  edges  ;    (£)  intersecting  edges  :   (^ 
edges  neither  parallel  nor  intersecting. 

8.  Examine,  in  the  same  way,  the  edges  of  the  prism  and  the  pyramid. 

9.  Which  edges  of  the  room  are  parallel  to  the  floor? 

10.  Give  another  instance  of  parallel  lines;   of  parallel  planes  ;   of  a  line 
parallel  to  a  plane. 

11.  Are  two  vertical  lines  parallel  to  each  other? 

Ans.  —  Strictly  speaking,  No;  because,  owing  to  the  round  shape  of  the  earth, 
two  vertical  lines  intersect  at  a  point  near  the  centre  of  the  earth.  But  the  dis- 
tance to  the  earth's  centre  (nearly  four  thousand  miles)  is  so  great  compared  with 
the  distance  apart  of  two  vertical  lines,  such  as  the  edges  of  a  house,  etc.,  that  the 
latter  are  always  regarded  as  parallel. 

12.  Are  horizontal  lines  always  parallel  to  each  other? 

13.  Can  two  vertical  planes  intersect  each  other? 

14.  Can  two  horizontal  planes  intersect  each  other? 

15.  Give   an   instance  of  parallel  lines  which  are  vertical ;   horizontal ; 
inclined. 

§  30.  The  simplest  way  to  draw  lines  parallel  to  each  other 
(especially  when  several  are  desired)  is  illustrated  in  Fig.  26. 
Besides  a  ruler,  a  piece  of  wood  with  three  smooth,  straight  edges 
called  a  SQUARE,  is  needed.  Keep  the  ruler  at  rest,  slide  the 
square  along  the  edge  of  the  ruler,  and,  in  different  positions  of 
the  square,  draw  lines  along  its  edge  ;  these  lines  will  he  parallel 
to  each  other.  To  draw  a  line  through  the  point  J/,  parallel  to  the 


3'-] 


CHAFl'ER    II. STRAIGHT    LINES. 


29 


line  P  Q,  place  the  ruler  and  square  as  shown  in  the  figure,  then 

slide  the  square  along  till  its 

edge  just   touches   M \    the 

line  drawn  through  Mt  along 

the  edge  of  the  square,  will 

be  the  parallel  required. 

NOTE.  —  How  parallel  lines  are 
drawn  with  ruler  and  compasses  will 
be  explained  later.  (See  $  78). 

Exercises.  —  1.  Drawastraight 
line ;  and  then,  through  a  point  not 
on  the  line,  a  parallel  to  it.  Pig-  26. 

2.  Draw  a  straight  line,  and,  through  a  point  on  the  line,  a  parallel. 

3.  Draw  five  parallel  lines  about  equally  distant  from  each  other. 

4.  Draw  free-hand  several  parallel  lines.      Test  and  correct  them  with 
ruler  and  square. 

IV.— Directions  of  Three  Lines. 

§  31.   Three  straight  lines  in  the  same  plane  present,  as  regards 
direction,  four  cases,  — 

(a)  The  three  lines  are  all  parallel ; 

(b)  Two  of  the  lines  are  parallel ; 

(c)  No  two  lines  are  parallel,  and  all  intersect  in  one  point ; 
(//)  No  two  lines  are  parallel,  and  they  do  not  all  intersect  in  a 

point. 

These  cases  are  all  illustrated  in  Fig.  27.     When  necessary,  the 

\ 


\ 


(a) 


\ 


Fig.  27. 


lines  must  be  prolonged  until  they  intersect.      In  case   (//)   the 
lines  (prolonged,  if  necessary)  intersect  in  three  points,  and  com- 


30  GEOMETRY    FOR    BEGINNERS.  [§  32- 

pletely  enclose   a  portion  of  a  plane  surface,  which  is   called  a 
TRIANGLE. 

NOTE.  —  The  properties  of  the  triangle  will  be  studied  in  Chapter  IV. 

Exercises.  —  1.  Draw  any  three  lines,  and  then  prolong  them  (by  dotted 
lines)  till  they  intersect. 

2.  In  how  many  points  can  four  lines  intersect  one  another?   (a)  when  all 
four  lines  are  parallel  ?   (£)  when  three  are  parallel  ?   (r)  when  two  are  par- 
allel ?  (</)  when  no  two  are  parallel  ? 

3,  Examine  five  lines  in  the  same  way. 

V. — Length  of  a  Line. 

§  32.  The  straight  line  which  passes  through  two  points,  A  and 
B  (fig.  28) ,  is  shorter  than  any  other  line 
passing  through  the  points,  and  hence  its 
length  is  taken  as  the  DISTANCE  of  the 
points  from  each  other. 

We  all  learn  as  soon  as  we  begin  to 
- 28-  walk  and  to   move    about  from  place  to 

place,  that  a  straight  line  is  the  shortest  distance  between  two 
points. 

Illustration. — We  may  illustrate  this  truth  as  follows  :  Fasten 
one  end  of  a  string  to  a  nail  which  is  fixed  in  a  wall,  and  pass  the 
other  end  through  a  ring  also  attached  to  the  wall.  Now  pull  the 


Fig.  29. 

free  end  with  one  hand,  and  the  part  of  the  string  between  the 
wall  and  the  nail  will  diminish  in  length  until  it  is  straight ;  and 
then  it  cannot  be  made  any  shorter. 


§33-]  CHAPTER    II. STRAIGHT   LINES.  31 

Application.  —  Sign-painters,  carpenters,  etc.,  make  use  of 
this  property  when  they  wish  to  trace  straight  lines  on  wood.  They 
chalk  a  cord  and  stretch  it  tightly  between  the  points  through  which 
the  line  is  to  pass.  Then  seizing  the  cord  by  the  middle,  they 


Fig.  so. 


draw  it  away  a  little  from  the  wood  and  then  let  it  go.  It  springs 
back,  strikes  the  wood  a  sharp  blow,  and  leaves  on  it  a  white  trace, 
which  is  a  straight  line. 

§  33.  As  regards  length,  two  lines  must  be  either  equal  or 
unequal. 

Definition.  —  Two  straight  lines  are  EQUAL,  if,  when  laid  one 
upon  the  other,  their  ends  or  extremities  can  be  made  to  coincide. 
If  this  cannot  be  done  the  lines  are  UNEQUAL. 

Expression  of  Equality. — If  two  lines,  AB  and  CD,  are 

equal,  we  express  the  fact  thus  :  A  B  =  CD.  A B 

This  expression  is  called  an  EQUATION,  and  is  c —D 

read  :  the  line  AB\s>  equal  to  the  line  CD.  E F 

The  sign  =  is  called  the  SIGN  OF  EQUALITY.  Fig.  31. 

Expression  of  Inequality. — The  lines  AB  and  E  F  {Fig. 
ji)  are  unequal  lines.  AB  is  greater  than  E  F,  or  EF  is  less 
than-  A  B.  In  place  of  the  words  in  italics,  we  often  employ  the 
signs  >  and  <,  and  write  :  AB>EF,  and  EF<AB. 

Exercise.  —  Read    the    following:     a  =  l>;    c>a;    x<y;    AJ3<PQ; 


32  GEOMETRY   FOR   BEGINNERS.  [§  34. 

§  34.  In  order  to  test  the  equality  or  inequality  of  two  lines,  it 
is  not  necessary  to  place  one  line  upon  the  other,  as  the  definition 
of  the  last  section  might  seem  to  require.  We  find  it  usually  more 
convenient  to  compare  both  lines  with  a  third  line,  which  can  be 
readily  placed  upon  or  by  the  side  of  each  of  them. 

In  constructing  geometrical  figures  and  diagrams,  we  make  this 
comparison  (especially  when  great  accuracy  is  required)  by  means 
of  the  dividers  (compasses).  Suppose,  for  example,  that  we  wish 
to  test  whether  the  lines  A  B  and  CD  (Fig.  ji)  are  equal.  Open 
the  dividers,  and  place  one  point  on  A,  the  other  on  B ;  this  is 
called  taking  the  distance  A  B  between  the  points  of  the  dividers. 
Then,  taking  care  not  to  alter  the  opening  of  the  dividers,  place 
one  point  on  C  and  observe  whether  the  other  point  will  fall  on 
D;  if  it  does  fall  on  D,  we  know  that  CD  =  A  B. 

NOTE.  —  In  general,  lines  are  compared  in  length  by  measuring  them,  and  ex- 
pressing their  lengths  in  feet,  miles,  metres,  etc.,  as  will  be  explained  in  Parts  VII. 
and  VIII.  of  this  chapter. 

Exercises.  —  1.    Draw  two  lines,  and  then  test  whether  they  are  equal. 

2.  Draw  a  horizontal  line  ;   then  draw  a  vertical  line  equal  to  it. 

3.  Explain  how  to  make  a  line  equal  to  a  given  line. 

4.  Draw  a  line  equal  to  one  edge  of  this  cube. 

5.  Draw  lines  equal  to  the  different  edges  of  this  prism. 

6.  Draw  four  lines,  a,  b,  c,  d,  as  follows  :  a  =  b,  c  >  a,  d<  b. 

7.  Draw  any  four  lines,  m,  n,  o,  p  ;  then  test  their  lengths,  and  write  the 
results  with  the  proper  signs  (=,  or  >,  or  <). 

8.  Try  to  draw,  free-hand,  two  equal  lines,  one  horizontal,  the  other  ver- 
tical ;   then  test  their  equality. 

9.  Draw  two  lines,  each  longer  than  the  greatest  opening  of  the  dividers. 
How  can  you  test  whether  they  are  equal  or  not? 

§  35.  In  testing,  with  the  dividers,  the  lengths  of  two  lines,  the 
third  line  with  which  we  compare  each  of  them  is  the  distance  be- 
tween the  points  of  the  dividers.  We  assume  that  if  the  two  lines 
are  each  equal  to  this  distance,  they  must  be  equal  to  each  other. 
This  assumption  is  so  very  obvious  that  it  almost  escapes  notice. 


§  36.]  CHAPTER   II. STRAIGHT   LINES.  33 

If  two  lines  —  or  any  two  magnitudes  —  are  each  equal  to  a  third, 
of  course  they  are  equal  to  each  other.  In  fact,  tjis  is  a  truth  so 
simple  in  its  nature  that  it  cannot  be  made  clearer  by  trying  to 
prove  it.  Truths  of  this  kind  —  that  is  to  say,  setf-evitftnttniths  — 
are  called  AXIOMS. 

Besides  the  axiom  here  under  consideration,  there  are  four 
others  often  useful  in  Geometry.  We  shall  now  state  them  all,  and 
shall  refer  to  them  hereafter  as  here  numbered. 

Axioms.  —  I.  If  two  magnitudes  are  each  equal  to  a  third, 
they  are  equal  to  each  other. 

II.  If  equals  are  added  to  equals,  the  sums  are  equal. 

III.  If  equals  are  subtracted  from  equals,  the  remainders  are 
equal. 

IV.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 

V.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 

§  36.  The  length  of  a  line  (like  its  direction)  is  susceptible  of 
change.  We  may  suppose  it  either  to  increase  or  to  diminish. 

Prolong  the  line  AB  (Fig.  32}  to  any  point  C ;  then  the  line 

A  C  is  equal  to  the  sum  of  the  lines        < 

A  B  and  B  C ;  here  we  may  employ    A*~~  ~~ 

to  advantage  the  sign  +  of  addition, 

and  write,  — 

A  C=  AB  +  B  C. 

Conversely,  the  line  A  B  is  the  difference  between  the  lines  A  C 
and  B  C ;  or  we  may  use  the  sign  —  of  subtraction,  and  write, — 

AB  =  AC-BC. 

Prolong  a  line  A  B  (Fig.  33}  by  adding  to  it  repeatedly  (by 
means  of  the  dividers)  lines  B  C,  CD,  D E,  etc.,  each  equal  to 
A  B.  Then  A  C  is  twice  as  long  as  A  B,  A  D  is  three  times  as 
long,  A  E  four  times  as  long,  etc.  That  is,  we  have  multiplied  the 


34  GEOMETRY    FOR    BEGINNERS.  [§   36. 

line  A  B  by  2,  3,  4,  etc.  It  is  usual  to  omit  the  sign  x  of  multi- 
plication, and  write  the  results  thus  :  AC=2AB,AD=$AB, 
A  E  =  4  AB,  etc.  Also 


Conversely,  a  line  can  always  be  divided  into  equal  parts.  In 
the  above  line,  for  example,  A  B  is  one-half  of  A  C,  one  third  of 
A  D,  one-fourth  of  A  E,  etc.  ;  in  other  words,  — 


234  3 

These  results  may  be  summed  up  in  the  following  general  truth  : 
Lines,  like  numbers,  may  be  added,  subtracted,  multiplied^  and  di- 
vided. 

Remark.  —  Observe,  however,  that  we  have  only  multiplied  and 
divided  a  line  by  a  number,  and  that  the  result  has  always  been 
another  line.  If,  for  instance,  we  multiply  the  line  AB  by  the 
number  10,  we  simply  repeat  the  line  a  certain  number  of  times, 
and  the  result  is  the  line  A  L  ;  or,  writing  the  operation  in  the 
form  of  an  equation,  — 

10  x  AB  =  AL. 

In  the  converse  operation  of  division  there  are  two  cases  :  - 

(a)  We  may  divide  the  line  ALty  the  number  10,  obtaining 
as  the  quotient  the  line  A  B.     In  this  case  we  divide  the  line  into 
a  certain  number  of  equal  parts. 

(b)  Or,  we  may  divide  the  line  A  L  by  the  .line  AB,  and  now 
our  quotient  will  be  the   number  10.     In  this  case  we  find  how 
many  times  one  line  is  contained  in  the  other  line.     In  this  sense 
division  is  the  comparison,  as  regards  magnitude,  of  two  things  of 
the  same  kind,  and  the  quotient  is  their  relative  magnitude  or  ratio. 


§  37-]  CHAPTER   II. STRAIGHT    LINES.  35 

Exercises.  —  1.  In  Fig.  33,  what  line  is  equal  to  BD  +  DE?  toAE — 
A  D  ?  to  three  times  the  sum  of  A  B  and  B  C  ?  to  four  times  the  difference 
of  A  Cand  CD? 

NOTE.  —  Perform  the  following  exercises  with  ruler  and  dividers,  and  not  with 
the  aid  of  the  divisions  of  a  divided  rule. 

2.  Draw  two  lines  ;  then  draw  lines  equal  (a)  to  their  sum,  (3)  to  their 
difference. 

3.  Drw  A£  =  CD,  and  MN—PQ;  then  draw  lines  equal  to  A  B + 
MN,  CD  +  PQ,AB  —  MN,  CD  —  PQ.     Why  are  the  first  "two  of  these 
lines  equal?     Why  are  the  last  two  also  equal? 

4.  Draw  a  line  three  times  as  long  as  your  pencil. 

5.  Draw  a  line,  A  B,  and  prolong  it  to  a  point,  P,  so  that  AP=  $A  B. 

6.  Explain  how  to  repeat  or  multiply  a  line  any  number  of  times. 

7.  Draw  a  line  equal  in  length  to  the  sum  of  the  edges  which  bound  one 
face  of  this  cube. 

8.  Draw  a  line  ;  then  divide  it  into  two,  four,  and  eight  equal  parts. 
NOTE. — This  is  to  be  done  here  by  repeated  trial  with  the  dividers. 

9.  Divide  a  line,  free-hand,  into  five  equal  parts.     Test  and  correct  with 
the  dividers.     Repeat  this  exercise  several  times. 

VI.  — Ratio  of  Two  Lines. 

§  37.  Definitions.  —  I.  If  a  line  is  contained  in  another  line 
one  or  more  times  without  a  remainder,  the  first  line  is  called  a 
MEASURE,  or  ALIQUOT  PART,  of  the  second ;  and  the  second  line  is 
called  a  MULTIPLE  of  the  first. 

II.  A  COMMON  MEASURE  of  two  or  more  lines  is  a  line  which  is 
exactly  contained  in  each  of  them. 

III.  The  RATIO  of  two  numbers  is  the  quotient  obtained  by  di- 
viding one  number  by  the  other ;  in  other  words,  it  expresses  how 
often  one  of  the  numbers  will  contain  the  other. 

IV.  The  RATIO  of  frvo  lines  is  the  number  of  times  one  of  the 
lines  will  contain  the  other  ;  it  is  equal  to  the  ratio  of  the  numbers 
which  express  how  often  a  common  measure  is  contained  in  each 
of  the  lines. 

V.  Of  the  two  terms  of  a  ratio,  the  first  is  called  the  ANTECEDENT, 
the  second  is  called  the  CONSEQUENT. 


36  GEOMETRY    FOR    BEGINNERS.  [§  38. 

For  example  :  in  Fig.  34,  where  AB  =  B  C  =  CD  —  D  E  —  E  F, 
the  line  AB  is  a  measure  of  AB,  of  A  C,  of  CF,  in  short,  of 
every  line  represented  in  the  figure  ;  therefore  A  B  is  a  common 
measure  of  all  these  lines.  On  the  other  hand,  all  the  lines  in  the 
figure  are  multiples  of  A  B. 


Fig-  34- 

If  we  compare  the  two  lines  A  B  and  A  F,  we  see  that  their 
common  measure  A  B  is  contained  once  in  A  B  and  five  times  in 
A  F;  hence,  the  lines  A  B  and  A  F  are  to  each  other  as  the  num- 
bers i  and  5  ;  that  is,  they  have  the  ratio  1:5.  By  expressing  this 
ratio  as  a  fraction,  we  obtain  the  equation,  — 

AB  _  i  X  A  B  _  i 
~AF  ~  5  x  AB  ~  5  ' 

The  reciprocal  ratio  of  AF  to  AB  is  that  of  5  to  i,  or  simply  5. 

NOTE.  —  In  common  language,  instead  of  using  the  term  ratio,  we  say,  A  B  is 
one-fifth  as  long  as  AF;  or,  reciprocally,  A  F"\s  five  times  as  long  as  A  B. 

In  like  manner,  it  is  easy  to  see  from  Fig.  34  that  — 
The  lines  A  B  and  A  C  have  the  ratio  1:2; 


2:5; 

AE        "       "         3:4;  etc. 
In  the  form  of  fractions,  these  ratios  would  be  written,  — 

^J?_i-  ^_*_     •   BD  !-2  •  £^_1  •  etc 
A  C~  2'  Z^~7~     '  AF~  $  '  AE~  4' 

Exercise.  —  What  is  the  ratio  of  A  D  and  A  B  (Fig.  34)  ?  AD  and  A  C? 
A  D  and  AD?  A  D  and  A  E  ?  AD  and  A  F?  A  Cand  A  E?  A  E  and  .  /  F? 
Write  the  values  both  with  dots  (:  )  and  as  fractions.  In  these  ratios,  which 
term  is  the  antecedent?  which  the  consequent? 

§  38.  In  order  to  find  the  ratio  of  two  lines,  apply  the  shorter 
line  to  the  longer  line  as  often  as  possible  ;  if  the  shorter  line  is  a 


'§  3  8.]  CHAPTER   II. STRAIGHT    LINES.  37 

measure  of  the  longer, — if,  for  example,  it  is  contained  in  the  longer 
exactly  five  times,  —  then  i  :  5  is  the  ratio  of  the  shorter  line  to  the 
longer. 

If,  however,  the  shorter  line  is  not  a  measure  of  the  longer,  the 
problem  becomes  more  difficult.  After  we  have  applied  the  shorter 
line  to  the  other  line  as  many  times  as  possible,  there  is  a  remainder 
less  than  the  shorter  line.  For  example  :  in  the  case  of  the  lines 
MNand  PQ  (Fig.  35),  when  MNhas  been  applied  twice  from 
P  towards  Q,  there  remains  the  portion  R  Q  of  the  longer  line. 
To  meet  this  difficulty,  apply  R  Q  to  M ' N  as  often  as  possible ; 
then  apply  the  new  remainder  SjVto  R  Q,  which  contains  it  ex- 
actly twice. 

This  series  of  operations  is  precisely  like  the  process  of  finding 
the  greatest  common  divisor  of  two  numbers ;  it  is,  in  fact,  the 
method  of  finding  the  greatest  common  measure  of  two  lines. 


P  R 

Fig-  35- 

From  what  precedes  we  see  that  — 
JR  Q  =  2  SN; 

MN=  3  R  Q  +  SN=  6  SN  +  SN=  7 
P  Q  =  2MN+  RQ  =  I46W  +  2SN 

hence      P  Q  -  l6  SN-  l6  •  and  MN      ^  SN 

--->       [= 


S  N  is  the  greatest  common  measure  of  the  two  lines  ;  the 
longer  contains  it  sixteen  times,  the  other  seven  times  ;  the  ratio 
of  the  lines,  therefore,  is  that  of  the  numbers  16  and  7,  either 
16  :  7  or  7  :  16. 


38  GEOMETRY    FOR    BEGINNERS.  [§  38. 

It  may  happen  that  we  cannot  find  a  common  measure  of  two 
lines  by  this  process,  for  the  reason  that  the  lines  have  no  common 
measure  ;  such  lines  are  called  INCOMMENSURABLE  LINES. 

Remark.  —  The  exact  ratio  of  two  incommensurable  lines  can- 
not be  expressed  by  numbers  ;  but  by  dividing  one  of  the  lines 
into  a  great  number  of  equal  parts,  and  applying  one  of  these  parts 
to  the  other  line  as  often  as  possible,  and  neglecting  the  remainder 
(which  will  be  very  small  compared  with  either  of  the  lines)  we 
are  able  to  approximate  very  nearly  to  the  true  value  of  the  ratio. 

If,  for  instance,  we  divide  one  of  two  lines  into  1000  equal 
parts,  and  find  that  the  other  line  contains  more  than  600,  but  less 
than  60  1,  of  these  parts,  we  know  that  the  true  value  of  the  ratio 
lies  between  $$#,  and  $$j\j.  For  nearly  all  purposes  we  should 
assume  that  the  ratio  was  r6o°(y°o>  or  I  \  m  other  words,  we  should 
assume  that  one  line  was  three-fifths  as  long  as  the  other. 

Exercises.  —  1.  In  Fig.  36,  of  what  lines  is  MN  a  common  measure? 
How  many  times  is  it  contained  in  each  of  them?  Of  what  lines  is  PQ 
(=±MN)  a  common  measure?  How  many  times  is  it  contained  in  each? 

2.    Express  the  values  of  the  other  lines  in  terms  of  P  Q. 


D 


Fig.  36. 

3.  How  many  times  is  i  P  Q  contained  in  each  line  in  Fig.  36  ?    Is  it  a 
common  measure  of  all  of  them?  If  we  divide  P  Q  into  three,  four,  five,  etc., 
equal  parts,  will  one  of  the  parts  be  a  common  measure  of  all  the  lines? 

4.  Of  what  lines  is  2  PQ  a  common  measure?  2  MN? 

5.  What  is  the  greatest  common  measure  of  MN  and  A  B?  of  A  B  and 
CD?  of  MN,  ABvn&CD?  of  MN,  A  B,  CD  and 


§  39-]  CHAPTER   II. STRAIGHT    LINES.  39 

6.  Find  the  ratio  of  P  Q  to  each  of  the  other  lines  in  Fig.  36,  and  write 
the  results  both  fractionally  and  with  dots. 

7.  Find  the  ratio  of  M ' N  to  all  the  other  lines,  and  write   the  results 
both  fractionally  and  with  dots. 

8.  Find  the  ratio  of  A  B  to  the  other  lines,  and  write  out  the  results. 

9.  What  is  the  ratio  of  CD  to  E  F?     What  is  the  reciprocal  ratio? 

10.  Draw  any  two  lines,  and  then  find  their  ratio. 

VII.  —  Units  of  Length. 

§  39.  In  comparing  the  lengths  of  lines  with  one  another  a  great 
amount  of  time  and  trouble  is  saved  by  employing  units  of  length. 

Definitions.  —  LA  UNIT  of  length  is  a  known  length  which 
men  agree  to  employ  as  a  common  term  of  comparison,  or  common 
measure  for  all  lines. 

Units  of  length  receive  individual  names,  as  the  foot,  the  yard, 
the  meter,  the  mile,  etc. 

11.  To  MEASURE  a  line  is  to  find  how  often  it  will  contain  a 
unit  of  length. 

In  other  words,  to  measure  a  line  is  to  find  its  ratio  to  the  unit 
of  length. 

III.  The  number  of  times  a  line  contains  a  unit  of  length 
is  the  NUMERICAL  VALUE  of  the  line  with  reference  to  that  unit; 
joined  to  the  name  of  the  unit,  the  whole  expression  is  called  the 
LENGTH  of  the  line. 

Examples. — 6  feet,  9  meters,  12  miles. 

NOTE.  —  In  olden  times,  units  of  length  were  usually  taken  from  some  part  of 
the  human  body.  The  word  foot  indicates  its  origin.  The  cubit  was  the  length  of  the 
forearm,  from  the  elbow  to  the  end  of  the  middle  finger.  The  first  attempt  to  es- 
tablish a  standard  yard  in  England  was  in  1120,  in  the  reign  of  King  Henry  I.,  who 
ordered  that  the  yard,  the  ancient  ell  (Latin  ulna,  elbow,  arm),  should  be  of  the 
exact  length  of  his  own  arm.  At  the  present  day,  horses  are  described  as  so  many 
" hands  high,"  and  a  place  is  sometimes  said  to  be  so  many  "paces  distant."  The 
span,  the  palm,  and  \htfathom  all  derive  their  origin  from  the  human  body;  the 
fathom  was  the  distance  between  the  extreme  points  of  the  outstretched  arms.  But 
the  units  of  length  now  recognized  by  law  among  civilized  nations  are  defined 
by  the  State  with  the  utmost  precision,  by  reference  to  a  standard  made  of  plati- 
num or  other  hard  metal,  which  suffers  no  sensible  change  from  age  to  age.  This 
standard  is  kept  at  the  capital  of  the  nation,  and  carefully  made  copies  of  it  ar? 
preserved  in  public  buildings  in  the  chief  cities  where  the  unit  is  employed. 


40  GEOMETRY    FOR    BEGINNERS.  [§  40 

§  40.  The  units  of  length  established  by  law,  and  used  for  mosl 
purposes  in  the  United  States  and  in  Great  Britain,  are  the  INCH, 
the  FOOT,  the  YARD,  the  ROD,  the  MILE. 

12  inches  (in.)  =  i  foot  (ft.).  5^  yards  =  i  rod. 

3  feet  =  i  yard  (yd.).         320    rods    =  i  mile. 

NOTE.  —  The  geographical  or  sea  mile  is  one-sixtieth  of  a  degree  of  a  meridian 
of  the  earth  and  is  equal  to  1.15  common  or  statute  miles,  very  nearly. 

Exercise.  —  How  many  inches  in  one  yard  ?  in  one  rod  ?  in  one  mile  ? 

§  41.  The  units  of  length  now  employed  by  the  people  of  most 
civilized  nations,  and  by  the  men  of  science  of  all  nations,  are 
called  METRIC  units,  from  the  name  of  the  fundamental  unit,  the 
meter.1  They  form,  taken  together,  a  part  of  the  Metric  System 
of  Weights  and  Measures,  first  adopted  by  the  French  just  after 
the  great  revolution  of  1792. 

Definition.  —  The  METER  —  the  fundamental  unit  from  which 
all  other  units  in  the  metric  system  are  derived —  is  defined  by  law 
as  the  length  of  a  certain  platinum  bar  kept  at  Paris. 

NOTES.  —  i.  The  nations  which  have  adopted  the  metric  system  take  from 
the  Paris  meter  the  most  perfect  copies  which  can  be  made,  and  in  the  capital  of 
each  nation  are  preserved  a  set  of  metric  measures  and  weights  with  which  those 
used  by  the  people  must  agree. 

2.  The  founders  of  the  metric  system  defined  the  meter  as  one  ten-millionth 
part  of  the  distance  from  the  Equator  to  the  North  Pole.  Extraordinary  efforts  were 
made  to  construct  a  platinum  bar  which  should  have  this  exact  value.  An  arc  of 
a  meridian  passing  through  France  and  Spain  was  measured  with  extreme  care, 
and  by  combining  the  results  with  other  known  facts  the  distance  from  the  Equator 
to  the  Pole  was  calculated.  The  platinum  bar  based  on  these  results  was  supposed 
at  the  time  to  be  exactly  what  it  was  denned  to  be.  But  not  long  afterwards  errors 
in  the  measurements  were  discovered,  and  it  is  now  known  that  the  Paris  meter 
(the  platinum  bar)  is  less  than  its  intended  value  by  an  amount  not  exceeding  — ^ 
of  the  whole. 

The  other  units  of  length  are  multiples  and  divisions  of  the 
meter  as  follows  :  — 

Ten  meters  are  called  one  DEKAMETER. 

One  hundred  meters  are  called  one  HEKTOMETER. 

1  The  word  meter  comes  from  a  Greek  word  meaning  measure,  and  is  pro- 
nounced me6-ter.  The  French  write  the  word  mttre. 


§  41.]  CHAPTER    II. STRAIGHT    LINES.  41 

One  thousand  meters  are  called  one  KILOMETER. 
One  tenth  of  a  meter  is  called  one  DECIMETER. 
One  hundredth  of  a  meter  is  called  a  CENTIMETER. 
One  thousandth  of  a  meter  is  called  a  MILLIMETER. 

NOTES. —  i.  Ten  thousand  meters  is  also  called  a  MYRIAMETER.. 

2.  The  names  of  the  multiples  de&ameter,  hectometer,  kilometer,  myn'ameter  are 
formed  by  prefixing  to  the  word  meter  the  Greek  numerals  deka  (io)P  hecto  (100), 
kilo  (1000),  and  myria  (10000)  ;  for  the  divisions  of  the  metre,  the  Latin  numep 
als,  deci,  centi,  and  milli  are  used. 

Abbreviations.  —  Meter  =  m  ;  dekameter  =  dkm  ;  hecto- 
meter =  hm  ;  kilometer  =  km  ;  decimeter  =  dm  ;  centimeter  =* 
cm  ;  millimeter  —  mm. 

TABULAR   VIEW. 

km.         hm.           dkm.                m.                   din.                     cm.  mm. 

I  =  10  =  IOO  =  IOOO  =  IOOOO  =IOOOOO  =  IOOOOOO 

I  =   10  =   IOO  =   IOOO  =   IOOOO  =  IOOOOO 

I  =    IO  =    IOO  =    IOOO  =  IOOOO 

I  =     10  =     IOO  =  IOOO 

I '  =       IO  =  IOO 

I  —  IO 

This  table  shows  that  a  given  number  of  units  of  one  value  are 
reduced  to  units  of  the  next  lower  value  by  multiplying  by  10,  and 
to  units  of  the  next  higher  value  by  dividing  by  10. 

Examples.  —  1.    Reduce  32  meters  to  decimeters,  and  also  to  dekameters. 

Ans.  32Odm;   3.2dkm- 

2.  Reduce  0.764  of  a  kilometer  to  meters. 

Solution.  —  Since  we  wish  to  reduce  kilometers  to  the  third  lower  unit,  we  must 
multiply  by  10  three  times.  As  in  decimal  fractions,  this  is  done  by  moving  the 
decimal  point  three  places  to  the  right.  Ans.  764™- 

3.  Reduce  376  millimeters  to  kilometers. 

Solution.  —  Here  we  have  to  reduce  millimeters  to  the  sixth  higher  unit;  there- 
fore, we  must  divide  by  10  six  times.  That  is,  move  the  decimal  point  six  places 
to  the  left.  Ans.  0.000376^", 

Remarks.  —  i.  The  units  most  used  are,  the  kilometer  (for 
long  distances),  the  meter,  and  the  centimeter  and  millimeter  (for 
lengths  less  than  one  meter) . 


42  GEOMETRY  FOR  BEGINNERS.  [§41. 

2.  In  practice,  only  one  unit  is  employed  to  express  a  length. 
We  say  76°"  or   76omm   (not  ydm  6cm)  ;    332™   (not  3hm    3dkm  2m, 
and  not  33dkm  2m),  etc. 

3.  In  reading  infraction  of  a  unit,  the  word  signifying  the  num- 
ber of  decimal  places  may  be  omitted ;  thus,  4.5™  may  be  read  : 
four  meters  five;    4.5 2m  may  be  read  :  four  meters  fifty-two,  etc. 
This  is  analogous  to  the  common  way  of  reading  money;    $2.75, 
for  instance,  is  read  :  two  dollars  seventy-five   (or,  yet  more  sim- 
ply :  two  seventy-five) .     In  the  case  of  money,  the  omitted  word 
is  cents ;  with  metric  units  it  is  tenths,  hundredths,  thousandths, 
etc.,  according  to  the  number  of  decimal  places. 

4.  In  adding,  subtracting,  etc.,  metric  units,  they  must  first  all 
be  reduced  to  a  common  unit,  as,  for  example,  the  meter.   Thus,  — 

6.4m  -f-  27cm  =  6.4m  f  0.27™  =  6.67m- 

COMPARISON   OF   METRIC   AND    ENGLISH    UNITS. 

kilometer  =  0.62137  miles  (about  |- of  a  mile), 

meter          =  1.0936    yards  (about  39-3-  inches), 

centimeter  =  0.3937    inch  (about  f-  of  an  inch), 

millimeter  =  0.0394    inch  (about  ^V  of  an  inch). 

mile  =1609  meters  (about  if- kilometers), 

yard  =       0.9144  meter  (about  -JT  meter), 

foot  =       0.3048  meter  (about  30  centimeters). 

inch  =        2.54  centimeters  (about  2^  centimeters). 

NOTE.  —  It  might  be  well  for  the  pupil  to  learn  by  heart  the  approximate  values 
(not  the  others)  just  given  in  parentheses;  but  the  best  way  to  become  familiar 
with  the  values  of  the  meter  and  its  divisions  is  by  comparing  a  meterstick  with 
a  yardstick,  by  comparing  the  divisions  on  a  divided  rule,  and  by  solving  exercises 
like  the  following. 

Exercises. —  1.  Give  the  metric  units  of  lengths  thus:  10  millimeters 
make  I  centimeter,  10  centimeters  make  I  decimeter,  etc. 

2.  Write  a  tabular  view  of  the  metric  units  of  length. 

3.  Write  with  abreviations :  8  meters;   72  kilometers;  324.6  dekameters; 
$7.34  hectometers;  9.27  centimeters;  4.2  decimeters;    13  millimeters. 


§  4I-]  CHAPTER   II.  —  STRAIGHT   LINES.  43 

4.  Read    the    following  :    37.5™  ;     947™™  ;    846.7cm  ;    l8dm ;     9.O4km; 
45.2o6hm  ;   77-96dkm. 

5.  Reduce  SS44mm  to  meters  and  to  kitometers. 

6.  Reduce  3-27km  to  meters  and  to  millimeters. 

7.  Reduce  I2.4hm  to  all  the  lower  units. 

8.  Reduce  i8ocm  to  all  the  higher  units. 

9.  What  part  of  a  kilometer  is  im  ?    icm  ?    imm  ? 

10.  25cm  are  what  part  of  a  meter?  also  what  part  of  a  kilometer? 

11.  24.32m  +  6dm  4-  i8dm  -f-  26ocm  =  how  many  meters? 

12.  4dkm  4-  4m  4-  4dm  4"  4cm  4-  4mm  =  how  many  meters? 

13.  6km  4-  6hm  4-  6dkm  4-  6m  4-  6dm  =  how  many  kilometers? 

14.  A  shopkeeper  sells  ribbon  as  follows  :    to  the  first  customer  8m,  to 
the  second  7OCm,  to  the  third  4.5™,  to  the  fourth  2dm,  to  the  fifth  8ocm;   how 
many  meters  does  he  sell  in  all? 

15.  What  is- the  difference  in  length  of  two  nails,  one  of  which  is  4cm,  the 
other  28mm  long? 

16.  From  iom  take  iomm. 

17.  From  iookm  take  ioomm. 

18.  87cm  X  36  =  how  many  meters? 

19.  42.64m  X  80  =  how  many  kilometers? 

20.  If  the  post-office  is  900™  from  my  house,  and  I  go  to  it  twice  a  day 
for  letters,  how  many  kilometers  do  I  walk  in  a  week? 

21.  I  receive  an  order  for  20  rulers,  each  to  be  3Ocm  long;   what  length 
of  wood  is  required? 

22.  How  many  rails  each  7.5m  long  are  required  to  build  a  track  3ookm 
long? 

23.  How  long  would  it  take  an  express  train  going  at  the  average  rate  of 
4Okm  an  hour  to  travel  from  Boston  to  San  Francisco,  a  distance  of  55OOkm? 

24.  If  im  of  silk  costs  $3.00,  what  costs  idm?  icm?  imm  ?  idkm?  ihm?  ikm? 

25.  Reduce  the  following  values  to  English  equivalents  :  velocity  of  an  ex- 
press train  =  6okm  an  hour,  height  of  Mt.  Blanc  =  481 5™,  usual  height  of 
a  barometer  =  j6cm. 

26.  Reduce  the  following  values  to  metric  equivalents :  velocity  of  light 
=  190,000  miles  a  second;  distance  from  Boston  to  New  York  =  236  miles; 
average  height  of  a  man  =  5  feet  8  inches. 

27.  Which  is  greater,  3km  or  2  miles?  3dm  or  I  foot? 

28.  What  unit  would  you  use  to  express  the  distance  from  New  York  to 
Chicago?  the  height  of  a  mountain?  the  length  of  a  room?  the  dimensions 
of  a  sheet  of  paper?  the  thickness  of  a  soap-bubble? 


44  GEOMETRY    FOR    BEGINNERS.  [§42. 

29.  Why  is  the  Metric  system  also  called  a  decimal  system? 

30.  Which  do  you  consider  preferable,  metric  units  or  the  foot,  yard,  etc.? 

VIII.  —  Measuring  a  Line. 

§  42.  In  order  to  measure  a  line  various  instruments  are  em- 
ployed. Among  them  we  may  mention  the  Divided  Rule,  the 
Yardstick  (or  Meterstick),  the  Tape-Rule  or  Roulette,  and  the 
Surveyor's  Chain. 

Lines  on  paper  are  generally  measured  by  placing  beside  them 
a  divided  rule.  A  very  useful  kind  of  divided  rule  is  a  rule  one 
foot  long,  divided  on  one  side  into  inches  and  parts  of  an  inch, 
on  the  other  into  decimeters,  centimeters,  and  millimeters.  On 
the  blackboard  a  yardstick  or  meterstick  is  usually  employed. 

Longer  lines,  such  as  the  length  of  a  room  or  of  a  garden,  can 
be  more  rapidly  measured  by  the  help  of  a  tape-rule  or  roulttte. 
These  are  made  of  cloth  or  flexible  steel,  wound  around  an  axis, 
and  enclosed  in  a  small  case  suitable  for  the  pocket.  They  are 
made  of  various  lengths  up  to  one  hundred  feet. 

In  measuring  lines  on  the  ground  surveyors  employ  a  chain 
sixty-six  feet  long,  containing  one  hundred  links,  or  a  steel  tape  one 
hundred  feet  long.  French  surveyors  use  a  chain  one  dekameter 
long,  containing  fifty  links.  If  a  chain  or  steel  tape  is  not  at  hand, 
a  rope  divided  by  knots  into  yards  or  meters  may  be  employed. 

In  the  absence  of  any  better  means  of  measuring  a  line,  pacing 
must  be  resorted  to.  The  average  value  of  a  pace  must  be  found 
by  counting  how  many  paces  are  made  in  walking  a  certain  known 
distance,  as  ten  meters.  Pacing  is  not  a  very  exact  way  of  meas- 
uring a  line,  but  it  is  a  rapid  way,  and  is  always  at  our  disposal. 

In  many  cases  a  line  is  not  measured  directly,  but  its  length  is 
computed  from  lines  of  known  length  by  methods  which  will  be 
described  hereafter.  Indeed,  it  often  happens  that  the  direct 
measurement  of  a  line  is  impossible  ;  for  example,  the  breadth  of 
a  river,  the  height  of  a  mountain,  the  diameter  of  the  earth,  and,  in 


§  43 -J  CHAPTER    II. STRAIGHT   LINES.  45 

general,  the  distance  of  an  inaccessible  object  (as  the  sun  or  the 
moon) . 

Exercises.  —  1.    Make  a  line  3dm  long,  and  divide  it  into  centimeters. 

2.  Draw  lines  with  lengths  as  follows  :   i6cm;   o.6dm  ;    I75mm. 

3.  Draw  lines  with  lengths  as  follows  :  j6cm;    I4dm;    i.36m. 

4.  Draw,  free-hand,  a  line  3dm  long.     Measure  it  with  the  divided  rule, 
and  find  what  error  you  have  made.     Repeat  the  exercise  three  times. 

5.  Measure,  in  metric  units,  with  the  divided  rule,  the  following  lines,  and 
write  out  the  results:    («)   Length,  breadth,  and  thickness  of  a  book;    (<£) 
length  and  breadth  of  your  desk;  (<r)  length  of  your  lead-pencil ;    (</)  length 
of  your  middle  finger. 

6.  Estimate  first,  by  the  eye,  then  measure  with  the  meterstick,  the  height 
of  the  door. 

7.  Measure,  with  a  roulette,  the  length  and  breadth  of  the  room;  also, 
the  length  and  breadth  of  the  play-ground. 

8.  Find  the  value  of  your  pace  by  pacing  twenty  meters. 

9.  Measure,  by  pacing,  the  distance  from  the  schoolhouse  to  your  home. 

§  43.  If  we  look  on  a  map  of  a  large  country  like  the  United 
States,  we  see  that  distances  which  amount  to  hundreds  of  miles 
are  reduced  in  length  and  represented  by  lines  at  most  a  few  inches 
long.  It  is  clear  that,  in  constructing  maps,  plans 1  of  estates,  etc., 
the  actual  lengths  of  lines  must  be  very  much  reduced  ;  for,  other- 
wise, the  map  or  plan  would  need  to  be  as  large"  as  the  country  or 
the  estate  which  it  represents. 

A  divided  line,  or  other  means  for  enabling  us  to  reduce  lengths 
measured  on  the  ground  in  a  given  constant  ratio,  is  called  a  RE- 
DUCING SCALE  ;  and  the  operation  of  drawing  on  paper  lines  whose 
length  shall  be  one-half,  one-fourth,  one-tenth,  or  any  other  frac- 
tional part  of  the  lines  measured  on  the  ground,  is  called  DRAWING 
TO  SCALE. 

1  Every  one  knows  what  a  map  is ;  a  plan  is  a  portion  of  a  horizontal  surface, 
or  surface  assumed  as  horizontal,  represented  on  paper  together  with  its  subdivis- 
ions and  other  details  of  importance  which  it  contains.  A  diagram  is  any  set  of 
lines  drawn  on  paper,  etc.,  to  illustrate  a  statement  or  aid  in  a  demonstration ;  the 
so-called  Figures  of  Geometry  are  diagrams. 


46  GEOMETRY   FOR   BEGINNERS.  [§43- 

Suppose,  for  example,  that  we  take  as  a  reducing  scale  the  di- 
visions and  subdivisions  of  a  divided  rule,  and  that  lines  are  to  be 
drawn  to  the  scale  of  i  meter  to  500  meters ;  in  other  words,  that 
the  ratio  or  scale  of  reduction  is  i  :  500.  Then  i  meter  would  be 
represented  on  the  paper  by  a  line  -^V"  —  2mm  l°ng>  a  distance 
of  48  meters  by  a  line  -±^m  =  9.6cm  long  ;  a  length  of  66.5  meters 
by  a  line  |^(fm  =  13.3°'"  long,  etc.  And,  in  general,  the  reduced 
length  would  be  found  by  dividing  the  real  length  by  500. 

Conversely,  the  real  length  of  a  line  would  be  found  from  the 
reduced  length  by  multiplying  the  reduced  length  by  500.  Thus, 
&  line  8  centimeters  long  on  the  paper,  would  represent  a  line  which 
was  in  reality  8cm  X  500  =  4ooocm  —  40™  long. 

Scales  are  named  by  expressing  the  ratio  of  reduction,  either 
fractionally,  thus,  T^<j,  -5-^,  TOIJFU-J  etc.,  or  with  two  dots,  thus, 
i  :  100,  i  :  500,  i  :  10000,  etc. 

A  reducing  scale  is  sometimes  printed  or  drawn  by  the  side  of  the 
map  or  diagram,  and  its  divisions  numbered  in  terms  of  the  real 
lengths  of  the  lines  represented  on  the  paper.  We  can  then  easily 
ascertain  with  the  dividers  the  real  length  represented  by  a  line  on 
the  paper.  How? 

The  scale  of  reduction  nmst  be  the  same  for  all  lines  on  the 
same  map,  plan,  or  diagram.  If  this  rule  were  not  followed,  the 
appearance  of  two  lines  on  the  paper  would  give  a  false  idea  of 
their  actual  relative  lengths  ;  and,  besides,  the  shape  of  the  map  or 
diagram  would  differ  from  the  shape  of  what  it  professed  to  rep- 
resent ;  in  other  words,  the  map  or  diagram  would  be  distorted. 

NOTES.  —  i.  The  choice  of  a  scale  depends  on  the  real  lengths  of  the  lines 
compared  with  the  size  of  the  paper  on  which  they  are  to  be  represented.  The 
scale  chosen  should  be  such  as  will  enable  us  to  represent  all  the  lines  on  the  paper, 
and  also  leave  a  fair  margin  around  the  edge  of  the  paper. 

2.  The  following  are  examples  of  scales  actually  employed:  —  United-States 
Coast  Survey,  charts  of  small  harbors,  etc.,  i :  5000  down  to  i :  60000 ;  charts  of 
bays,  sounds,  etc.,  1:80000  down  to  1:200000;  general  charts,  usually,  1:400000; 
Ordnance  Survey  of  Great  Britain,  partly  i  inch  to  I  mile  (i :  63360),  and  partly  6 
inches  to  i  mile  (i :  10560) ;  Government  Survey  of  France,  original  scale,  i :  20000. 
scale  of  copies,  i :  40000  and  i :  80000. 
3.  For  ordinary  purposes,  a  reducing  scale  need  not  be  made  on  the  paper ;  we 


§  43-]  CHAPTER   II. STRAIGHT    LINES.  47 

choose  the  scale,  then  compute  the  reduced  lengths,  and  then  lay  off  these  lengths 
on  the  paper  with  the  aid  of  a  divided  rule.  But  when  great  accuracy  is  demanded, 
it  is  necessary  to  have  on  the  paper  a  reducing  scale  by  means  of  which  we  can  de- 
termine the  length  of  a  line  with  extreme  precision.  Such  a  reducing  scale  is  de- 
scribed in  Chapter  VIII.  (See  p.  i84-) 

Exercises.  —  1.   Draw  five  parallel  lines,  and  take  on  them  to  the  scale 
I  :  200  the  distances  30™,  2Om,  9m,  48™,  I5m. 

2.  Draw  any  three  lines,  and  find  what  lengths  they  represent  to  the  scale 
I  :  200. 

3.  What  is  the  value  of  the  scale  if  im  is  represented  by  4cm  ? 

4.  If  5cm   represent  ioom  natural  length,  what  length  will  represent  im? 

2Q»«  ?    jdkm  ?    i2^m  p 

5.  In  what  ratio  are  natural  lengths  reduced  if  the  scale  is  i  inch  to  the 
foot?  i  inch  to  the  mile?    2  millimeters  to  the  kilometer? 

Ans.  i :  48  ;    i :  126720 ;    1 :  500000. 

6.  The  distance  between  two  towns,  drawn  on  paper  to  the  scale  i :  40000, 
is  represented  by  a  line  idm  long;   find  the  distance  between  the  towns. 

Ans.  4km. 

7.  In  a  railroad  survey  the  total  length  is  eighty  miles.     What  scale  will 
enable  us  to  represent  a  plan  of  the  road  on  a  piece  of  paper  just  one  foot 
in  length? 

8.  Draw  on  the  blackboard  three  horizontal  lines.     Then  represent  them 
on  paper,  choosing  a  suitable  scale  for  this  purpose.     Write  the  value  of  the 
scale  on  the  paper  under  the  lines. 


REVIEW   OF   CHAPTER   II. 
QUESTIONS. 

1.  How  many  directions  has  a  straight  line? 

2.  How  is  a  straight  line  determined  in  position? 

3.  How  is  a  straight  line  drawn? 

4.  What  is  free-hand  drawing? 
5*  What  is  a  phimb  line? 

6.  Define  vertical,  horizontal,  and  inclined  lines,  and  give  examples. 

7.  Define  vertical,  horizontal,  and  inclined  planes,  and  give  examples. 

8.  How  can  you  test  whether  a  line  or  a  plane  is  vertical? 
9»  How  can  you  test  whether  a  line  or  a  plane  is  horizontal? 


48  GEOMETRY    FOR    BEGINNERS. 

10.  When  is  a  line  said  to  rotate  about  a  point?     When  does  the  rotation 
amount  to  one  revolution  ? 

11.  Define  a  circumference ;  \\s,  centre  ;  an  arc ;  a  radius;  a  circle. 

12.  What  is  true  of  radii  of  the  same  circle? 

13.  Describe  how  circles  (arcs,  etc.)  are  drawn. 

14.  In  considering  the  directions  of  two  lines,  what  two  cases  arise? 

15.  Define  parallel  lines,  and  give  examples. 

16.  Show  that  parallel  lines  can  never  meet. 

17.  Are  lines  which  would  never  meet  always  parallel?     Give  an  example. 

18.  Give  examples  of  parallel  curves ;  of  parallel  planes  ;  of  a  line  and  a 
plane  parallel  to  each  other. 

19.  How  are  parallel  lines  drawn? 

20.  What  four  cases  arise  when  we  consider  the  directions  of  three  lines? 
Represent  the  four  cases  by  figures. 

21.  How  can  we  illustrate  the  truth  that  a  straight  line  is  the  shortest  dis- 
tance between  two  points?     Give  an  application  of  the  same  truth. 

22.  Define  equal  and  unequal  lines. 

23.  Explain  how  the  equality  and  the  inequality  of  two  lines  are  expressed. 

24.  How  is  the  equality  of  two  lines  tested? 

25.  What  is  an  Axiom  ?    Give  Axioms  I.,  II.,  III.,  IV.  and  V. 

26.  Explain  how  lines  are  added,  subtracted,  multiplied,  and  divided. 

27.  In  what  two  senses  of  the  word  can  division  be  performed  on  a  line? 

28.  Define  a1 measure  of  a  line;   a  multiple  of  a  line;   a  common  measure  of 
two  or  more  lines;   the  ratio  of  two  numbers  and  of  two  lines.     What 
are  the  terms  of  a  ratio  called? 

29.  In  what  two  ways  can  a  ratio  be  expressed  or  written? 

30.  How  is  the  ratio  of  two  lines  found? 

31.  What  are  incommensurable  lines? 

32.  How  can  we  approximate  to  the  value  of  their  ratio? 

33.  Define  a  unit  of  length;  measuring  a  line;  the  numerical  value  and  the 
length  of  a  line. 

34.  What  is  the  ratio  of  a  foot  to  an  inch?  of  a  yard  to  a  foot?  of  a  rod  to  a 
yard?  of  a  mile  to  afoot?    What  are  the  corresponding  reciprocal  ratios? 

35.  Define  the  meter. 

36.  Name  and  define  the  multiples  and  divisions  of  the  meter. 

37.  How  are  metric  units  reduced  from  one  denomination  to  another? 

38.  Which  of  the  metric  units  are  most  used? 

39.  In  expressing  a  length,  how  many  units  should  be  used? 

40.  How  would  you  read  a  length  containing  a  fraction  of  a  unit? 


CHAPTER   II. STRAIGHT   LINES.  49 

41.-  In  adding,  subtracting,  etc.,  metric  units,  what  rule  must  be  observed? 

42.  What  is  the  value  of  one  meter  in  inches? 

43.  How  are  lines  on  paper  usually  measured?  longer  lines,  such  as  the 
length  of  a  room?  lines  on  the  ground? 

44.  What  are  the  advantages  and  disadvantages  of  pacing  as  a  means  of 
measuring  a  line? 

45.  Give  examples  of  lines  which  cannot  be  directly  measured. 

46.  What  is  a  reducing  scale,  and  what  is  drawing  to  scale  ? 

47.  I  low  are  scales  of  reduction  named  and  written? 

48.  Why  must  the  ratio  or  scale  of  reduction  be  constant  for  all  lines  belong- 
ing to  the  same  map*  plan,  or  diagram  ? 


EXERCISES. 

1.  How  many  straight  lines  can  be  drawn  through  the  North  and  South 
poles  of  the  earth? 

2.  How  many  straight  lines  can  be  drawn  through  n  points  (it  being  any 
number),  each  line  connecting  two  points? 

Solution. —  From  the  first  point  we  can  draw  lines  through  all  the  other  points; 
therefore,  in  all,  n — I  lines.  The  same  number  may  be  drawn  from  each  of  the 
other  points,  and  since  there  are  n  points,  this  would  make  n  (n — i)  lines  in  all. 
But,  in  this  way,  two  lines  would  be  drawn  in  every  case  through  the  same  two 
points ;  these  two  lines,  however,  would  coincide  and  form  but  one  line  by  $  22,  II. 
Therefore,  the  total  number  of  different  straight  lines  which  can  be  drawn  is 
n  (n—  i) 

2, 

3.  Find  the  greatest  possible  number  of  straight  lines  which  can  be  drawn 
through  ten  points.  Ans.  45. 

4.  Through  a  vertical  line  how  many  vertical  planes  can  be  passed  in 
thought  (that  is,  how  many  different  vertical  planes  can  be  imagined, 
all  of    which    shall    contain  the   vertical   line)  ?    how  many  horizontal 
planes?   how  many  inclined  planes? 

5.  Through  a  horizontal  line   how  many  vertical  planes  can  be  passed? 
how  many  horizontal  planes?  how  many  inclined  planes? 

6.  Through  an  inclined  line  how  many  vertical  planes  can  be  passed?  how 
many  horizontal  planes?  how  many  inclined  planes? 

7.  How  many  centres  can  a  circle  have?    how  many  circles  can  have  the 
same  centre? 

8.  Can  two  curved  surfaces  be  parallel  to  each  other?  a  curved  surface  and 
a  plane  surface? 


50  .          GEOMETRY    FOR    BEGINNERS. 

9.    Can  you  draw  a  line  parallel  to  a  given  line  with  the  aid  only  of  the 
dividers  ? 

10.  Find  in  how  many  points  six  lines  can  intersect  one  another. 

11.  Can  you  illustrate  with  numbers  Axioms  II.-V.? 

12.  If  you  multiply  the  ratio  of  two  lines  by  their  reciprocal  ratio,  what  is 
the  product  always  equal  to? 

13.  Draw  any  two  lines,  and  find  their  ratio. 

14.  Choose  two  lines  in  the  room  ;   estimate  their  ratio  by  the  eye,  then  find 
it  by  measuring  them. 

15.  i8m  -r-4dm—  64cm-|-  268mm  =  how  many  meters? 

16.  9km  —  220m  -4-  84dkm  —  72  hm  =  how  many  kilometers ? 

17.  The  sum  of  two  lines  is  24  feet,  and  their  difference  is  9  feet ;   find  the 
lines. 

18.  Two  lines  are  to  each  other  as  5  : 3,  and  their  sum  is  48m;   find  the  lines. 

Ans.  i  X  48m=  30™,  and  f  X  48  m  —  i8m. 

19.  The  sum  of  two  lines  is  84°™,  and  their  ratio  is  5:16;   find  the  lines. 

20.  The  difference  of  two  lines  is  20™,  and  their  ratio  is  4:9;   find  the  lines. 

Ans.    |  X  20m  =  i6m,  and  |  X  20™  =  36™. 

21.  I  read  in  a  newspaper  that  "the  river  Theiss  has  fallen  3ocm";   how 
much  is  this  in  inches? 

22.  Among  the  scales  prescribed  for  the  United-States  engineer  service  are  : 
general  plans  of  buildings,  I  inch  to   10  feet;   maps  i-i  miles  square,  2 
feet  to  i  mile  ;   do.,  3  miles  square,  I  foot  to   I  mile  ;   do.,  from  4  to  8 
miles  square,  6  inches  to  I  mile  ;   do.,  9  miles  square,  4  inches  to  I  mile  ; 
do.,  not  exceeding  24  miles  square,  2  inches  to  I  mile  ;   do.,   50  miles 
square,  I  inch  to  I  mile.     Express  these  scales  as  fractions,  with  unity 
for  the  numerator. 

Ans.  T10-,  2FIOJ    52lgO>   TU5F<J>  T5&4'S>   3T68tf>  F3&60- 

23.  Which  scale  reduces  lines  the  most :  a  scale  of  I  inch  to  i  mile,  or  a  scale 
of  I  centimeter  to  I  kilometer? 

24.  If  the  scale  to  which  a  map  is  made  is  not  known,  how  can  it  be  found? 


§440 


CHAPTER   III.  —  ANGLES. 


51 


CHAPTER  III. 
ANGLES. 

CONTENTS.  —  I. .Definition  of  an  Angle  (§44).  II.  Magnitude  of  an  Angle 
(§§45,46).  III.  Magnitudes  of  Particular  Angles  (§§  47-49).  IV.  Measure 
of  an  Angle  (§$  50-52).  V.  Angles  made  by  Two  Lines  (§§  53,  54).  VI. 
Angles  made  by  Three  Lines  (§§  55-58). 


L  —  Definition  of  an  Angle. 

§  44.  When  we  open  the  legs  of  the  compasses  (Fig.  J/)  they 
point  in  different  directions  :  they 
are  now  said  to"  make  an  ANGLE 
with  each  other.  Replace,  in 
thought,  the  legs  by  geometrical 
lines  A  B  and  A  C,  and  we  have 
an  angle  as  understood  in  Geome- 
try. 

Definition.  —  An  ANGLE  is  the 
difference  in  direction  of  two  lines. 

The  two  lines  are  called  the  SIDES  of  the  angle,  and  the  point 
where  they  meet  is  called  the  VERTEX.!  If  the  lines  do  not 
actually  meet  (see  Fig.  25,  p.  27),  the  point  where  they  would 
meet  if  prolonged,  is  the  vertex  of  the  angle. 

An  angle  is  named  in  one  of  three  ways :  — 

(1)  By  a  small  letter  placed  just  inside  the  vertex; 

(2)  By  a  large  letter  placed  just  outside  the  vertex  ; 

(3)  By  three  letters,  one  at  the  vertex,  the  others  on  the  sides. 


1  Plural,  vertices. 


52  GEOMETRY    FOR    BEGINNERS.  [§45- 

In  the  last  case,  in  reading  or  writing  the  angle,  the  letter  at 
the  vertex  must  stand  between  the  others. 

For  example  :  the  angle  of  the  lines  A  B  and  A  C  (Fig.  37)  is 
called  either  the  angle  d,  or  the  angle  A,  or  the  angle  B  A  C  or 
CAB.  In  general,  the  word  angle  need  not  be  written  ;  the  let- 
ters d,  or  A,  or  £  A  C,  or  CAB  are  sufficient. 

/  Exercises.  — 1.    Name  the  angles  drawn  on  the  black- 

/  board. 

2.    Draw  four  angles  ;   name  them  in  different  ways. 
3*    If  (as   in  Fig.  j<?)   several  angles  have  the  same 
p.       r,  vertex,  which    method   of   naming   an  angle    cannot   be 

used?     Why  not? 


II. — Magnitude  of  an  Angle. 

§  45.  Angles  differ  in  magnitude.  We  shall  get  a  clearer  no- 
tion of  the  magnitude  of  an  angle  if  we  regard  the  angle  as  gen- 
erated or  described  by  the  rotation  of  a  line. 

Hold  one  leg  of  the  compasses  at  rest,  and  make  the  other  rotate 
about  the  pivot  in  the  direction  shown  by  the  arrow  (Fig.  37)  ; 
then  the  angle  made  by  the  leg  will  increase  in  magnitude.  On 
the  other  hand,  if  we  rotate  the  legs  in  the  opposite  direction,  the 
angle  will  diminish ;  and  when  the  legs 'are  brought  together,  it  will 
vanish  entirely. 

In  like  manner,  we  may  regard  the  angle  made  by  any  two  lines, 
A  B  and  A  C  (fig.  3f)9  as  described  by  the  rotation  of  one  of  the 
lines  about  their  intersection  A,  until  it  coincides  in  direction  with 
the  other  line. 

It  is  obvious  that  the  size  or  magnitude  of  an  angle  depends 
entirely  on  the  amount  of  rotation  required  to  describe  it,  and  not 
at  all  on  the  lengths  of  its  sides. 

Definition. —  Two  angles  are  EQUAL  if  they  can  be  placed  one 
upon  the  other  so  that  their  vertices  coincide  in  position,  and  their 
sides  coincide  in  direction  ;  otherwise  the  angles  arc  UNEQUAL. 


§46.] 


CHAPTER   III.  — ANGLES. 


53 


We  shall  soon  see  that  the  equality  of  two  angles  (like  that  of 
two  lines)  can  be  tested  without  actually  placing  one  upon  the 
other. 

Exercises.  —  1.  Of  the  two  angles  on  the  blackboard,  which  has  the 
longer  sides?  which  is  the  greater  angle? 

2.    Must  the  two  sides  of  an  angle  have  the  same  length? 


§  46.  If,  in  the  angle  B  s±  ^  ^  ./.  jp),  the 
side  A  C  is  made  to  rotate  about  the  vertex 
A  away  from  A  B,  until  it  comes  into  the 
position  A D,  it  is  clear  that  the  angle  BAD 
is  equal  to  the  sum  of  the  angles  B  A  C  and 
CAD:  that  is, — 

BAD=BA  C+  CAD. 


39- 


•D 


If,  on  the  other  hand,  the  side  AD  of  the  angle  BAD  be 
made  to  rotate  towards  A  B  until  it 
arrives  at  the  position  A  C,  there  re- 
mains the  angle  B  A  C,  which  is  equal 
to  the  difference  of  the  angles  BAD 
and  CAD:  in  other  words, — 

BAC=BAD-  CAD. 

If  the  angles  AOB,BOC,C  OD, 
D  OE,  E  OF  (Fig.  40),  are  equal 
to  one  another,  then  A  OC  =  2  A  OB, 
AOD=$AOB,  A  OE=  4 A  OB,AOF= 

Conversely,  it  is  evident  that,  — 

A  OB  =  $A  O  C=  ±A  OD  =  ±A  OE  =  I A  OF. 

Angles,  in  the  same  sense  as  lines,  may  be  added,  subtracted, 
multiplied,  and  divided. 

Definitions.  —  I.  To  BISECT  an  angle  (or  a  line)  is  to  divide  it 
into  two  equal  parts. 

II.  A  line  which  bisects  an  angle  is  called  a  BISECTOR. 


54 


GEOMETRY    FOR    BEGINNERS. 


[§47- 


Exercises.  —  1.   In  Fig.  40  what  angle  is  equal  to  the  sum B  0  C-\-  COE? 
to  the  difference  A  O  F—  C  O  F? 

2.  Draw  freehand,  three  angles,  of  which  one  is  twice,  the  other  five  times 
as  large  as  the  first. 

3.  Bisect  freehand,  a  given  angle. 

4.  How  many  bisectors  can  an  angle  have? 

5.  Divide  an  angle,  freehand,  into  three,  four,  five,  six  equal  parts. 

6.  In  Fig.  40  what  is  the  ratio  of  the  angles  A  O  B  and  A  OF?  A  0  C 
and  A  OE? 


III.  —  Magnitudes  of  Particular  Angles. 

§  47o   Angles  receive  different  names  according  to  their  mag- 
nitude. 

If  we  conceive  a  line  to  revolve 
round  O  (Fig.  41) ,  starting  from  the 
initial  position  O  A  and  revolving  in 
the  direction  of  the  arrow,  it  will 
describe  an  angle  of  constantly  in- 
creasing magnitude. 

When  the  line  has  made  exactly 
one-fourth  of  a  complete  revolution, 
the  angle  described,  A  O  C,  is  called 
a  RIGHT  ANGLE  ;  it  is  sometimes  de- 
noted by  the  letter  R. 


Fig.  41. 


When  the  line  has  made  one-half  of  a  revolution,  it  is  in  the 
position  OE ;  the  angle  described,  A  OE,  is  equal  to  two  right 
angles,  and  is  called  a  STRAIGHT  ANGLE,  because  its  sides  O  A  and 
O  E,  having  opposite  directions,  form  one  straight  line. 

When  the  line  has  made  three-fourths  of  a  revolution,  it  is  in 
the  position  O  G.  The  angle  which  has  been  described,  A  O  G, 
is  equal  to  three  right  angles. 

Finally,  when  the  line  has  returned  to  the  initial  position  O  A, 
it  has  made  a  complete  revolution,  and  has  described  four  right 
angles. 


§  47-]  CHAPTER   III. ANGLES.  55 

Definitions.  —  I.  A  STRAIGHT  ANGLE  is  the  angle  described  by 
a  line  which  makes  one -half  of  a  revolution. 

II.  An    angle   less  than  a  straight  angle  is  called  a  CONCAVE 
ANGLE  ;  an  angle  greater  than  a  straight  angle  is  called  a  CONVEX 
ANGLE. 

III.  A  RIGHT  ANGLE  is  one-half  of  a  straight  angle. 

IV.  A  concave  angle,  if  less  than  a  right  angle,  is  called  an 
ACUTE  ANGLE  ;  if  greater,  it  is  called  an  OBTUSE  ANGLE. 

In  Fig.  41,  A  OB,  A  O  C,  A  O D  are  concave  angles ;  A  OF, 
A  O  G,  A  O  H  are  convex  angles ;  A  O  B  is  an  acute,  A  O  D  an 
obtuse  angle. 

NOTES.  —  i.  Acute  and  obtuse  angles  are  sometimes  called  oblique  angles,  in 
distinction  from  the  right  angle. 

2.  Whenever  two  lines  meet,  there  is  formed  both  a  concave  and  a  convex  angle ; 
but,  when  not  otherwise  stated,  the  concave  angle  is  to  be  understood. 

Since  the  sides  of  a  straight  angle  form  a  straight  line,  and  one 
straight  line  can  always  be  placed  on  another  straight  line  so  as  to 
coincide  with  it  in  direction,  it  follows  that  two  straight  angles  sat- 
isfy the  test  of  equal  angles  (§  45)  ;  in  other  words,  — 

All  straight  angles  are  equal  to  one  another. 

And  since  a  right  angle  is  half  of  a  straight  angle,  it  follows 
(by  Axiom  III.)  that  — 

All  right  angles  are  equal  to  one  another. 

Exercises.  —  1.  Open  the  legs  of  the  compasses  so  that  they  make  aright 
angle ;  a  straight  angle  ;  three  right  angles  j  an  acute  angle ;  an  obtuse 
angle  ;  a  concave  an'gle  ;  a  convex  angle. 

2.   On  the  blackboard  are  several  angles  :  what  kind  is  each,  and  why? 

3»    What  kind  of  angles  are  found  on  the  cube?  the  prism?   the  pyramid? 

4.  Point  out  angles  on  various  objects  (doors,  windows,  tables,  walls  of 
the  room,  etc.),  and  name  the  kind  in  each  case.     What  kind  of  angle  occurs 
most  frequently? 

5.  Draw,  free-hand,  a  right  angle;    an  acute  angle;    an  obtuse  angle;    a 
concave  angle;   a  convex  angle. " 

6.  At  12  o'clock  what  angle  do  the  hands  of  a  watch  make  with  one  an- 
other? at  3  o'clock?  at  6  o'clock?  at  9  o'clock. 


56 


GEOMETRY    FOR    BEGINNERS. 


[§48. 


7.  Mention  a  time  of  the  day  when  the  angle  between  the  hands  of  a 
watch  is  an  acute  angle?  a  right  angle?  an  obtuse  angle? 

8.  What  part  of  one  revolution  does  the  minute-hand  describe  in  5  min- 
utes?   (A MS.  -/Q  =iV-)     In  I0>  20>  3°>  40.  5°  minutes? 

9.  How  long  does  it  take  the  minute-hand  to  describe  a  right  angle?  the 
hour-hand? 

10.  What  kind  of  an  angle  does  the  hour-hand  describe  in  I  hour?    in  3 
hours?  in  5  hours?  in  6  hours?  in  8  hours?  in  9  hours?  in  12  hours? 

11.  What  kind  of  an  angle  does  a  vertical  line  make  with  a  horizontal 
line? 

12.  Draw  a  right  angle  with  one  side  vertical ;    with  one  side  horizontal; 
with  one  side  inclined. 


§  48.   Definitions.  —  I.  A  straight  line  is  said  to  be  PERPEN 
DICULAR  to  another  straight  line  when  it  makes  a  right  angle  with  z>, 

II.   Two  lines  not  perpendicular  to  each  other  are  said  to  be 
INCLINED  to  each  other. 

Examples. — The  lines  AE  and  C  G  (Fig.  41},  the  sign  -f 

in  arithmetic,  a  common  cross, 
the  slats  of  a  window  (Fig. 
42) ,  are  examples  of  perpen- 
dicular lines ;  the  lines  A  B 
and  A  C  (Fig.  39)  are  inclined 
lines. 

The  sign  1  is  sometimes  used 
for  the  word  perpendicular; 
thus  :  A  EL  CG  is  read  :  the 
line  A  E  is  perpendicular  to  the 
line  C  G. 

Exercises.  —  1.  Give  examples 
of  perpendicular  lines  ;  of  inclined 
lines. 

2.    What  is   the    difference   be- 
tween  a  perpendicular  line  and  a 
Fig.  42. '  vertical  line  ? 


§  49-]  CHAPTER   III. ANGLES.  57 

3.  If  two  lines  are  perpendicular  to  each  other,  and  one  of  them  is  ver- 
tical, must  the  other  be  horizontal? 

4.  If  two  lines  are  perpendicular  to  each  other,  and  one  of  them  is  hori- 
zontal, must  the  other  be  vertical  ? 

5.  Give  an  example  of  two  perpendiculars,  one  of  which  is  vertical,  the 
other  horizontal. 

§  49.  Among  the  most  common  problems  which  the  surveyor, 
carpenter,  draughtsman,  etc.,  have  to  solve,  are  — 

(/.)  To  erect  a.  perpendicular  at  a  given  point  of  a  given  line. 

(//.)  To  let  fall  a  perpendicular  from  a  given  point  to  a  given  line. 

On  the  ground,  surveyors  use  various  instruments,  the  simplest 
of  which  is  the  Surveyors'  Cross  {Fig.  43). 

It  consists  of  a  block  of  wood  having  two  saw-cuts  made  very 
precisely  at  right  angles  to  each  other.  This  block 
is  fixed  on  a  pointed  staff  on  which  it  can  turn 
freely.  To  erect  a  perpendicular,  set  the  cross  at 
the  point  of  a  line  where  a  perpendicular  is  wanted, 
then  turn  the  block  till  on  looking  through  one  saw- 
cut  you  see  the  ends  of  the  line ;  then  the  other 
saw-cut  will  point  out  the  direction  of  the  perpen- 
dicular. To  let  fall  a  perpendicular  to  a  line  from 
some  object  (as  the  corner  of  a  field,  a  tree,  etc.), 
set  the  cross  at  a  point  of  the  line  which  seems  to  i&'  43~ 

the  eye  about  right,  then  note  how  far  from  the  object  the  perpen- 
dicular at  this  point  strikes,  and  move  the  cross  that  distance ;  re- 
peat this  operation  till  the  correct  point  is  found. 

On  paper,  or  the  blackboard,  both  of  the  above  problems  are 
readily  solved  by  means  of  the  Ruler  and  the  Square.  Explain 
how. 

NOTE. —  The  method  of  solving  these  problems  on  paper  with  the  compasses 
will  be  shown  in  Chapter  IV. 

Exercises.  —  1.  Draw  a  straight  line,  and  then,  free-hand,  (ci)  erect  a 
perpendicular  at  a  point  on  the  line,  (£)  let  fall  a  perpendicular  from  a  point 
not  on  the  line.  Test  and  correct  the  results  by  the  aid  of  the  ruler  and 
the  square. 


58  GEOMETRY    FOR    BEGINNERS.  [§   50. 

2.  How  many  perpendiculars  can  "be  erected  at  a  given  point  of  a  given 
line?     How  many  can  be  let  fall  from  a  given  point  to  a  given  line? 

3.  Draw  a  horizontal  line,  and,  at  five  points  on  it,  erect  perpendiculars. 
What  kind  of  lines  are  they? 

4.  Draw  a  vertical  line,  and,  at  five   points  on  it,  erect  perpendiculars. 
What  kind  of  lines  are  they? 

5.  Draw  an  inclined  line,  and,  at  five  points  on  it,  erect  perpendiculars. 
What  kind  of  lines  are  they? 

6.  Draw  a  horizontal  line,  erect  a  perpendicular,  and  at  a  point  of  this 
perpendicular  erect  also  a  perpendicular.     What  direction  has  the  second  per- 
pendicular? 

7.  Erect  a  perpendicular  at  a  point  of  a  line  on  the  ground. 

8.  Choose  a  point  in  the  yard  (playground  or  field),  and  let  fall  a  perpen- 
dicular to  the  nearest  side  of  the  yard. 

IV. — Measure  of  an  Angle. 

§  50.   Angles  (like  lines)  are  measured  by  choosing  a  unit,  and 
then  comparing  other  angles  with  this  unit.     The  unit  for  measur- 
ing angles  is  obtained  by  dividing  the  right  angle  (which  would  be 
too  large  a  unit)  into  ninety  equal  parts,  and  calling  one  of  these 
parts  a  degree.     The  degree  is  subdivided  into  minutes  and  seconds. 
One  ninetieth  part  of  a  right  angle  is  called  a  DEGREE. 
One  sixtieth  part  of  a  degree  is  called  a  MINUTE. 
One  sixtieth  part  of  a  minute  is  called  a  SECOND. 
Abbreviations.  —  Degree  =  °  ;  minute  =  ' ;  second  =  ". 
Thus,  24°  1 7'  8"  is  read :  24  degrees  1 7  minutes  8  seconds. 
The  values  in  degrees  of  the  angles  defined  in  §  47  are  as  fol- 
lows :  — 

A  right  angle          =    90°.  An  acute  angle    <    90°. 

A  straight  angle      =180°.  An  obtuse  angle  >    90°. 

Three  right  angles  —  2  70°.  A  concave  angle  <  1 80°. 

Four  right  angles   =  360°.  A  convex  angle  >  180°. 

Exercises. — 1.    How  many  seconds  in  i°?  in  io°?  in  45°?  in  360°? 

2.  Reduce  48°  54'  36"  to  seconds. 

3.  Reduce  120000"  to  degrees. 


§51-] 


CHAPTER    III. ANGLES. 


59 


4.  Add  37°  48'  35",  28°  39',  and  78°  9'  55". 

5.  From  128°  15'  31"  take  69°  42'  18". 

6.  Multiply  1 8°  35'  by  2,  3,  4,  and  5. 

7.  Divide  72°  27'  by  2,  3,  4,  and  5. 

8.  Reduce  to  degrees  5  R  (5  right  angles);    6  R  ;    8  A1/    12  A5;    %  R  ; 
R;  %  R;  &  A'/  i  AV  i  R;  %  R;  \  R;  T\>  R;  TV  A. 

9.  How  many  right  angles  in  540°?  900°?  225°?  30°?  5°? 
10.  Find  the  ratio  between  10°  and  three  right  angles. 


§  51.  In  measuring  angles  we  take  advantage  of  a  simple  rela- 
tion between  angles  and  the  arcs  intercepted  between  their  sides, 
and  described  with  the  same  radius  from  their  vertices  as  centres. 

In  Fig.  44  compare  the  angles  A  O  B,  A  O  C,  A  O  D,  with  the 
arcs  A  B,  AC,  AD,  intercepted  by 
their  sides.  We  see  that  the  greater 
the  angle  the  greater  the  corresponding 
arc,  or  arc  intercepted  by  its  sides. 
This  conclusion  is  general :  in  the 
same  circle,  the  greater  the  angle  at  the 
centre  the  greater  the  corresponding 
arc,  and  the  less  the  angle  the  less  the 
arc. 

Again,  if  the  two  angles  A  OB  and 
FOG  are  equal,  the  corresponding 
arcs  A  B  and  FG  are  also  equal.  For  the  angles,  being  equal, 
may  be  placed  one  on  the  other  so  that  they  will  coincide ;  and 
then  the  arcs  A  B  and  F  G  must  also  coincide,  since  all  the  points 
of  one  arc  are  at  the  same  distance  from  the  centre  O  as  the  points 
of  the  other  arc. 

In  like  manner  it  can  be  shown  that  if  the  arcs  A  B  and  F  G 
are  equal,  the  corresponding  angles  at  the  centre,  AOB  and  FOG, 
will  also  be  equal. 

The  same  reasoning  may  be  applied  with  the  same  result  to  any 
angles  and  their  corresponding  arcs  ;  hence,  in  general :  — 


Fig.  44. 


60  GEOMETRY   FOR    BEGINNERS.  [§  51. 

I.  —  Eqtial  angles  at  the  centre  of  a  circle  intercept  on  the  cir- 
cumference equal  arcs. 

II.  —  Equal  arcs   on  the   circumference    correspond  to    equal 
angles  at  the  centre. 

Accordingly,  the  circumference  of  a  circle  is  divided  into  360 
equal  arcs,  each  corresponding  to  an  angle  of  i°  at  the  centre. 
These  arcs  are  also  called  degrees,  and  are  subdivided  like  the  angu- 
lar degree  into  minutes  and  seconds.  And,  in  practice,  an  angle  is 
measured  by  finding  how  many  degrees,  minutes,  and  seconds 
there  are  in  the  corresponding  arc,  it  being  obvious  that  the  num- 
ber of  angular  degrees,  minutes,  and  seconds  in  an  angle  is  the 
same  as  the  number  of  arc  degrees,  minutes,  and  seconds  in  the 
arc  described  with  any  radius  from  the  vertex  of  the  angle  as  a 
centre,  and  intercepted  between  its  sides. 

We  have  just  said  that  the  arc  employed  to  measure  an  angle 

may  be  described  with  any  ra- 
dius. With  a  longer  radius  we 
have,  it  is  true,  a  longer  arc 
(Fig.  45),  but  each  division  of 
the  arc  is  also  correspondingly 
increased  in  length,  so  that  the 
number  of  divisions  (degrees, 
etc.)  in  the  entire  arc  remains 
the  same  as  before. 

We  are  now  prepared  to  an- 
swer the  question  :  How  can  the  equality  of  two  angles  be  tested 
without  placing  one  of  them  upon  the  other  ?  We  reply  :  two  an- 
gles are  equal,  if,  when  their  corresponding  arcs  are  measured  (as 
shown  in  the  next  section),  these  arcs  are  found  to  contain  the 
same  number  of  degrees,  minutes,  and  seconds. 

NOTE. — How  an  arc  can  be  divided  into  degrees,  etc.,  is  a  question  which  must 
be  deferred  until  we  know  more  about  the  properties  of  the  circle. 


§   52.]  CHAPTER   ill. ANGLES.  61 

Exercises.  —  1.  Find  the  angle  described  by  the  hour-hand  of  a  watch  in 
I  hour;  2  hours;  5  hours;  12  hours. 

2.  Find  the  angle  described  by  the  minute-hand  of  a  watch  in  i,  5,  10,  15, 
20,  30,  45  minutes. 

3.  In  how  many  minutes  does  the  minute-hand  describe  an  angle  of  i°? 
60° ?  225°?  300°? 

4.  What  angle  do  the  hands  of  a  watch  make  with  each  other  at  I  o'clock? 
at  2,  3,  etc.,  up  to  12  o'clock? 

5.  The  earth  turns  on  its  axis  in  24  hours  ;   what  angle  does  a  point  on  the 
surface  describe  in  I  hour?  in  6,  12,  15  hours?  52  hours? 

§  52.  On  paper  or  the  blackboard,  when  great  accuracy  is  not 
required,  angles  are  measured  with  an  instrument  called  a  PRO- 
TRACTOR (Fig.  46).  It  is  a  semicircle,  made  of  paper,  horn, 


Fig.  46. 

brass,  or  silver,  the  circular  edge  of  which  is  divided  into  180  equal 
parts  or  degrees.  To  measure  an  angle  with  the  protractor,  place 
the  centre  of  the  instrument  on  the  vertex  of  the  angle,  and  its 
zero  line  on  one  side  ;  then  read  off  on  the  edge  of  the  protractor 
the  division  through  which  the  second  side  of  the  angle  passes. 
On  the  ground,  surveyors  and  engineers  employ  for  measuring 


62 


GEOMETRY    FOR   BEGINNERS. 


[§52. 


angles,  costly  instruments  called  THEODOLITES.  A  cheap  substitute 
for  a  theodolite  is  shown  in  Fig.  4.7.  It  consists  of  two  pieces  of 
wood  shaped  like  rulers  mounted  on  a  vertical  axis,  by  a  pin  driven 


Fig.  47. 

through  their  exact  centres.  The  vertical  needles  inserted  near 
the  end  of  the  rulers  are  used  for  sighting.  In  place  of  the  needle 
nearest  the  eye,  it  is  better  to  employ  a  thin  strip  of  wood,  At 
having  a  fine  vertical  slit ;  and  in  place  of  the  other  needle,  a  ver- 
tical wire  fixed  in  a  light  frame,  B. 


By  the  help  of  this  instrument,  arid  a  protractor,  one  can  mea*. 
ure  with  considerable  accuracy  an  angle  on  the  ground  •  for  in- 
stance, the  angle  M ON  (Fig.  48). 


§  53-]  CHAPTER   III.  —  ANGLES.  63 

Exercises.  —  1.    On  the  blackboard  are   several   angles.     Measure  each 
with  the  protractor,  and  write  the  result  between  its  sides. 

2.  Draw  five  different  angles.     Estimate  the  value  of  the  first  in  degrees; 
then  measure  it  with  a  protractor.     Proceed  in  like  manner  with  each  of  the 
other  angles.     Give  the  results  in  3  columns:  in  column  I,  the  estimated  val- 
ues ;  in  column  2,  the  measured  values  ;   in  column  3,  the  differences  between 
the  estimated  and  the  measured  values. 

3.  From  a  point  on  a  straight  line  draw  two  lines,  both  on  the  same  side 
of  the  given  line  ;   measure  each  of  the  three  angles  thus  formed,  andadd  the 
results.     What  is  the  sum?     What  ought  the  sum  to  be? 

4.  Through  a  point  draw  three  lines  ;  measure  the  six  angles  thus  formed 
about  the  point  ;  add  the  results.     What  is  the  sum?     What  ought  it  to  be? 

5.  Open  the  legs  of  the  compasses  so  that  they  make  angles  of  90°;   60°; 

45°;  30°. 

6.  Imagine  two  lines  drawn  from  your  eye,  one  to  the  right-hand  upper 
corner  of  the  room,  the  other  to  the  left-hand  upper  corner  ;    what  angle 
would  the  lines  make? 

7  .    With  the  aid  of  the  protractor  make  an  angle  equal  to  a  given  angle  ? 

8.  Draw  angles  equal  to  20°;   30°;   40°;   60°;   90°;    100°;    150°;   180°; 

15°;  45°;  79°;  81°;  142°. 

9.  Explain  how  the  instrument  in  Fig.  4.7  is  constructed,   and  how  you 
would  proceed  to  measure  by  means  of  it  the  angle  M  O  N'\n  Fig.  48. 

V.  —  Angles  made  by  Two  Lines. 

§  53.   The  angles  A  O  Cand  C  OB  (Fig.  49)  are  called  adjacent 
angles. 

Definition  I.  —  Two  angles  are 
ADJACENT  when  they  have  the  vertex 
and  one  side  common,  and  the  other 
sides  are  opposite  parts  of  the  same 
straight  line.  _ 

We  see  from  the  figure  that  — 


Since  this  equation  would  hold  true  in  whatever  direction  the  line 
O  C  was  drawn,  we  conclude  that  in  all  cases  — 

The  sum  of  two  adjacent  angles  is  two  right  angles,  or  180°. 


64  GEOMETRY   FOR   BEGINNERS.  [§  54. 

Definition  II. — If  the  sum  of  two  angles   is    180°,  each  is 

called  the  SUPPLEMENT  of  the  other. 

Adjacent  angles  are  always  supplements  of  each  other. 

Exercises.  —  1.    If  one  of  two  adjacent  angles  is  a  right  angle,  what  is 
the  other? 

2.  If  one  of  two  adjacent  angles  is  acute,  what  is  the  other? 

3.  If  one  of  two  adjacent  angles  is  obtuse,  what  is  the  other? 

4.  If  one  of  two  adjacent  angles  is  known,  how  is  the  other  found? 

5.  How  is  the  supplement  of  a  given  angle  found? 

6.  Find  the  adjacent  angles  of  the  following  angles:    10°;   30°;   45°; 
75°;    99°;    100°;    179°;    15°  48';    79°  13'  52".     What  is  the  supplement  of 
each  of  these  angles? 


§  54.  Whenever  two  lines  cross  one  another  there  are  formed 
about  the  intersection  of  the  lines  four 
angles,  a,  b,  c,  d  (Fig.  jo).  Of  these 
angles  the  pairs  a  and  b,  b  and  c,  c  and  d> 
—  d  and  a  are  adjacent  angles  ;  the  pairs 
a  and  c9  b  and  d  are  called  vertical 
angles. 

Definition.  —  Two  angles  are  called 

Fig.  co. 

VERTICAL  angles  if  they  have  a  common 

vertex,  and  the  sides  of  one  angle  have  opposite  directions  to  the  sides 
of  the  other  angle. 

If  we  measure  two  vertical  angles,  as  a  and  c  (Fig.  50) ,  either 
with  a  protractor  or  by  cutting  them  out  on  a  piece  of  cardboard 
and  laying  them  one  upon  the  other,  we  shall  find  that  they  are 
equal.  But  without  measuring  them  we  can  prove  that  they  must 
be  equal,  if  we  bear  in  mind  what  has  been  said  about  adjacent 
angles  in  the  last  section.  Since  b  is  an  adjacent  angle  to  both 
a  and  c,  therefore, — 

and 


§   55-]  CHAPTER   III. ANGLES.  65 

Hence  it  follows  from  Axiom  I.  (state  the  axiom)  that,  — • 

a+b=b+c 
subtract  b  =  b 

There  remains  a  =  c 

For,  if  equals  are  taken  from  equals,  the  remainders  are  equal 
(Axiom  III.). 

Since  this  reasoning  holds  good,  however  the  intersecting  lines 
cut  each  other,  we  come  to  the  general  conclusion,  that  — 

Two  vertical  angles  are  always  equal  to  each  other. 

Remark.  —  In  the  above  reasoning  we  have  made  use  of  the 
general  truth  shown  in  the  last  section ;  namely,  that  the  sum  of 
two  adjacent  angles  is  180°  ;  and  also  of  Axioms  I.  and  III.  By 
reasoning  upon  these  truths  we  have  proved  or  demonstrated  a  new 
truth ;  namely,  that  two  vertical  angles  are  equal.  Geometrical 
truths  which  are  capable  of  proof  by  reasoning  from  known  truths 
are  called  THEOREMS. 

Exercises.  —  1.    Draw  adjacent  and  vertical  angles  and  name  them. 

2.  What  is  the  sum  of  the  angles  a  and  b  (Fig. 50)!     Why? 

3.  What  is  the  sum  of  the  angles  a,  b,  c,  and  d  (Fig.  jo~)  ?     Why? 

4.  One  of  the  angles  formed  by  two  intersecting  lines  is  36°;    find  the 
others. 

5.  One  of  the  angles  formed  by  two  intersecting  lines  is  90° ;   find  the  others. 

6.  Draw  any  two  intersecting   lines,  and  find  the  values   of  the   angles 
formed  at  the  intersection. 

?•  If  one  of  the  angles  formed  by  two  intersecting  lines  is  known,  how 
are  the  others  found? 

8.  Each  of  two  vertical  angles  is  45°;   find  the  value  of  each  angle  in  the 
other  pair  of  vertical  angles. 

9.  Prove  that  b  =  d  (Fig.  50)  by  reasoning  like  that  used  above  to  show 
that  a=c. 

VI.  —Angles  made  by  Three  Lines. 
§  55.   Thus  far  the  angles  considered  have  had  a  common  ver- 
tex j  let  us  now  consider  angles  formed  about  two  different  ver- 


66  GEOMETRY    FOR    BEGINNERS.  [§55- 

tices.     Such  angles  are  formed  when  two  straight  lines  are  cut  by 
a  third  line. 

The  most  important  case,  and  the  only  one  which  we  shall  exam- 
ine, is  that  in  which  two  parallel  lines, 
AB  and  CD  (Fig.  jz),  are  cut  by  a 
-%     third   line,  E  F.     There   are   formed 
about  the   two  points  of  intersection 
_     eight  angles.      These   angles   receive 
D     special  names. 

The  four  angles,  c,  d,  n>  m,  which 
lie  between  the  parallel  lines,  are  called 

INTERNAL  ANGLES  ;  the  other  four,  a,  b,  0,  /,  are  called  EXTERNAL 
ANGLES. 

An  external  angle  and  an  internal  angle,  as  a  and  m,  having  dif- 
ferent vertices,  and  lying  on  the  same  side  of  the  intersecting  line, 
are  called  CORRESPONDING  ANGLES.1 

An  external  angle  and  an  internal  angle,  as  a  and  «,  having  dif- 
ferent vertices,  and  lying  on  different  sides  of  the  intersecting  line, 
are  called  CONJUGATE  ANGLES. 

Two  external  angles,  as  a  and  p,  or  two  internal  angles,  as  d  and 
m,  having  different  vertices,  and  lying  on  the  same  side  of  the  in- 
tersecting line,  are  called  OPPOSITE  ANGLES. 

Two  external  angles,  as  a  and  o,  or  two  internal  angles,  as  d  and 
n,  having  different  vertices,  and  lying  on  different  sides  of  the  in- 
tersecting line,  are  called  ALTERNATE  ANGLES. 

Exercises.  —  1.  Name  the  four  pairs  of  corresponding  angles  in  Fig.  51 ; 
the  four  pairs  of  conjugate  angles ;  the  four  pairs  of  opposite  angles ;  the 
four  pairs  of  alternate  angles. 

2.  Which  are  external,  which  internal  opposite  angles?  which  are  ex- 
ternal, which  internal  alternate  angles? 

8»  Which  is  the  angle  corresponding  to  o?  conjugate  to  p?  opposite  to 
m  ?  alternate  to  n  ? 

— _ 1 _— . — __ . — — 

1  They  are  also  called  external-internal  angles. 


§  56.]  CHAPTER   III. ANGLES.  67 

4.  Draw  two  parallel  lines,  and  a  third  line  intersecting  them.  Letter  the 
eight  angles  e,f,g,  h,  and/,  r,  s,  t.  Then  write  in  six  vertical  columns  (a) 
the  external  angles  ;  (li)  the  internal  angles  ;  (c)  the  corresponding  angles; 
(c/)  the  conjugate  angles ;  (i)  the  opposite  angles ;  (/)  the  alternate 
angles. 

§  56.   If  we  conceive  the  line  AB  to  move  along  the  line  E  F, 
remaining   always  parallel  to  its   first 
position,  then,  since  its  direction  does 
not  change,  the  four  angles  a,  l>,  c,  d    ; 
which    it    makes   with   EF  do    not 
change.     When  A  B  reaches  the  par- 
allel line  CD,  it  will  coincide  with  it 
in  direction,  and  the  four  angles  a,  b, 
c,  d  will  coincide  respectively  with  the 

corresponding  angles  m,  ;/,  o,  p.  Moreover,  each  pair  of  conjugate 
angles,  as  a  and  ;/,  become  now  adjacent  angles ;  therefore  their 
sum  is  1 80°  (§  52)  :  the  same  is  true  of  each  pair  of  opposite  an- 
gles, as  a  and  p ;  lastly,  each  pair  of  alternate  angles,  as  a  and  o, 
become  now  vertical  angles,  so  that  they  are  equal  (§  53). 
Hence,  — 


I. 

2. 

3- 

4- 

a  =  m. 

a  +  n=  2  R. 

a  +  p  =  2  R. 

a=  o. 

b  =  n. 

b  +771=2  R. 

b  +  0=2  R. 

b  =  p. 

C  =0. 

c  +  /  =  2  R. 

c  -f-  n  —  2  R. 

c  =  m. 

d  =  p. 

d+  o  =  2  R. 

d-\-m=  2  R. 

d=  n. 

Putting  these  results  into  words,  we  obtain  the  following  theo- 
rem :  — 

Theorem.  — If  two  parallel  lines  are  cut  by  a  third  line,  — 
/.   The  corresponding  angles  are  equal ; 

2.  The  sum  of  two  conjugate  angles  is  two  right  angles  ; 

3.  TJic  sum  of  two  opposite  angles  is  two  right  angles  ; 
4..   The  alternate  angles  are  equal. 


68  GEOMETRY   FOR   BEGINNERS.  ['§  57. 

liemark.  —  Theorems   often  involve  consequences  which  are 

easily  deduced  from  the  theorems  : 
such   consequences    are    called    in 
_     Geometry  COROLLARIES.     The  pres- 
ent theorem  furnishes  an  example. 
Corollary.  —  If  a  =  90°  (that  is, 


D  ,toowstatw  =  90 

(that   is,  that    CD-lEF)  ;   or,  to 
Fig.  53.  state     the     corollary     in     general 

terms,  — 

If  a  straight  line  is  perpendicular  to  one  of  two  parallels,  it  is 
also  perpendicular  to  the  other. 

Exercises.  —  1.  Give  the  proof  separately  for  each  of  the  last  three 
(equations  in  each  of  the  above  four  series  of  equations. 

Models.    b  =  n,  because  when  A  B  coincides  with  CD,  b  coincides  with  n. 

b  -J-  m  =  2  R,  because  when  A  B  coincides  with  C  D,  b  becomes  adjacent  to  m, 
and  the  sum  of  two  adjacent  angles  is  180°. 

2.  Find  the  other  seven  angles  when  (7  =  112°;   when  a  =  90°. 

3.  Draw  two  parallel  lines,  and  a  line  intersecting  them;   then  find  the 
eight  angles  thus  formed,  using  a  protractor  as  little  as  possible. 

4.  If  either  one  of  the  four  parts  in  the  theorem  of  this  section  is  not 
true,  what  can  we  conclude  as  to  the  lines  A  B  and  CD? 

§  57.  Theorem.  —  If  two  straight  lines  are  cut  by  a  third  line 
.so  that  two  corresponding  angles  are  equal,  then  the  lines  must  be 
parallel. 

If,  for  example,  a  =  m  (Fig.  52),  then  is  AB  \\  CD.  For,  if 
we  move  A  B  towards  C  D,  keeping  it  parallel  to  its  first  position, 
during  the  motion  the  angle  a  does  not  change  because  the  direc- 
tion of  A  B  does  not  change.  When  the  intersection  of  A  B  and 
E  F  coincides  with  the  intersection  of  CD  and  F  F,  AB  must 
coincide  in  direction  with  CD,  since  a  =  m.  Therefore,  AB,  in 
its  first  position,  must  be  parallel  to  CD. 

Corollary.  —  By  making  in  this  theorem  a  =  m  =  90°,  we  ob- 
tain the  following  corollary  :  — 


§  58.]  CHAPTER   III. ANGLES.  69 

If  two  straight  lines  are  both  perpendicular  to  a  third  line,  they 
are  parallel.  (Fig.  54.) 

Remark.  —  All  the  four  properties  enumerated  in  the  theorem 
of  the  last  section  are  so  related  that  if  either  one  of  them  is 
true  the  other  three  must  also  be  true,  and  the  two  intersected 
lines  must  be  parallel. 

Hence  arises  the  importance  of  corresponding,  conjugate,  oppo- 
site, and  alternate  angles.  Before  we  can  assert  with  absolute  cer- 
tainty that  two  lines  are  parallel,  we  ought  to  show  that  the  lines, 
when  prolonged  farther  than  even  the  imagination  can  reach,  would 
not  meet.  Such  an  actual  extension  of  the  lines  P  R 
is  of  course  impossible  ;  it  is,  however,  quite  un- 
necessary, for  the  parallelism  of  two  lines  is  very 
simply  tested  by  the  angles  which  are  formed 
when  the  lines  are  cut  by  a  third  line. 

Exercises.  —  1.    Prove   the    corollary,   independently 
of   the  theorem,  with  lines  drawn   and  lettered  as   in 

&%•  54- 

2»    When  more  than  two  lines  are  perpendicular  to  a      r^ — TJ i — TV 

third,  are  they  all  parallel  to  each  other? 

Fig.  54- 

§  58.  In  Fig.  55,  we  have  drawn  AB  II  CD  and  EF  \\  G H. 
If  we  compare  the  angle  a  with  either  of  the  angles  x,  y,  *,  we  see 
that  their  sides  are  respec- 
tively parallel.  But  there 
are  these  differences  :  (Y.)the 
sides  of  x  have  the  same  di- 
rection  respectively  as  those 
of  a ;  (it.)  the  sides  of  z 
have  opposite  directions ;  -^ 
(///.)  the  sides  of y  have,  one 
the  same  direction,  the  other 
the  opposite  direction,  ~.  . 

fjs  •  55* 


70  GEOMETRY    FOR    BEGINNERS. 

In  case  (/.),  a  =  b  (corresponding  angles  on  the  parallels  EF 
and  G  H},  and  x  =  b  (corresponding  angles  on  the  parallels  C D~ 
and  A  B)  :  therefore  a  =  x  (Axiom  I.). 

Theorem  I.  If  two  angles  have  their  sides  respectively  par- 
allel, and  directed  the  same  way  from  the  vertex,  the  angles  are 
equal. 

In  case  («.),  x  =  z  (vertical  angles).  And  a  =  x  (Theorem  I.). 
Therefore  a  =  z  (Axiom  I.). 

Theorem  II.  If  two  angles  have  their  sides  respectively  par- 
allel, and  directed  opposite  ways  from  the  vertex,  the  angles  are  equal. 

In  case  (in.),  x  +  y  =  2R  (adjacent  angles).  But  a  =  x 
(Theorem  L).  Therefore  a  -f-  y  =  2  R. 

Theorem  III.  If  two  angles  have  their  sides  respectively  par- 
allel, and  directed,  the  one  pair  the  same  way,  the  other  pair  opposite 
ways ',  from  the  vertex,  the  sum  of  the  angles  is  two  right  angles. 

Exercises.  —  1.    In  Fig.  55,  if  a  =  115°,  what  is  the  value  of  x?  of  y  ?  ofz? 

2.  Prove  Theorem  I.,  using  in  the  proof  the  angle  c  in  Fig.  55,  instead  of 
the  angle  b. 

3.  What  relation  exists  between  a  and  s  {Fig.  55)  ?     Why? 

4.  Prove  Theorems  I.,  II.,  and  III.,  with  a  new  figure,  drawn  and  lettered 
differently  from  Fig.  33. 


REVIEW   OF   CHAPTER   III. 
QUESTIONS. 

1.  Define  an  angle;  its  sides ;  its  vertex. 

2.  How  is  an  angle  named? 

3.  How  may  an  angle  be  conceived  as  produced  by  motion? 

4.  On  what  does  the  magnitude  of  an  angle  depend? 

5.  Define  equal  angles. 

6.  Explain   (by  figures)  what  is  meant  by  adding,  subtracting,  multiplying, 
and  dividing  angles. 


CHAPTER    III. ANGLES.  71 

7  *    What  is  bisecting  an  angle  or  a  line  ?     What  is  the  bisector  of  an  angle  ? 
.    8.    Define  a  straight  angle  ;    a  concave  angle  ;     a  convex  angle  ;    a  right 

angle  ;   an  acute  angle  ;   an  obtuse  angle. 

9.    Show  that  all  straight  angles  are  equal ;   also,  that  all  right  angles  are 
equal. 

10.  When  are  two  lines  perpendicular  to  each  other?  when  are  they  inclined? 

11.  WThat  are  two  of  the  commonest  problems  in  surveying  and  drawing? 

12.  Describe  the  surveyor's  cross,  and  how  to  use  it. 

13.  How  are  perpendiculars  to  a  given  line  drawn  on  paper? 

14.  Define  a  degree,  a  minute,  and  a  second. 

15.  W7hat  are  the  values  in  degrees  of  the  angles  mentioned  in  Question  8? 

16.  Explain  why  an  angle  can  be  measured  by  means  of  the  arc  of  a  circle 
intercepted  between  its  sides. 

17.  Why  is  the  measure  of  an  angle  the  same  for  any  radius  of  the  arc? 

18.  How  is  the  equality  of  two  angles  tested  without  placing  one  of  them 
on  the  other? 

19.  Describe  the  protractor,  and  how  to  use  it. 

20.  Describe  a  simple  substitute  for  a  theodolite,  and  how  an  angle  on  the 
ground  is  measured  by  means  of  it. 

21.  Define  adjacent  angles  ;   the  supplement  of  an  angle. 

22.  Show  that  the  sum  of  two  adjacent  angles  must  be  180°. 

23.  Define  vertical  angles. 

24.  Two  vertical  angles  are  equal.1 

25.  What  is  a  theorem  ? 

26.  Define  external  angles  ;   internal  angles  ;   corresponding  angles  ;   conju- 
gate angles  ;   opposite  angles  ;   alternate  angles. 

27.  If  a  straight  line  intersect  two  parallel  lines,  the  corresponding  angles 
are  equal,  the  alternate  angles  are  equal,  and  the  sum  of  two  conjugate 
angles,  or  of  two  opposite  angles,  is  180°. 

28.  What  corollary  follows  from  this  theorem? 

29.  What  is  a  corollary  ? 

30.  If  two  lines  are  cut  by  a  third  line  so  that  two  corresponding  angles  are 
equal,  the  two  lines  are  parallel. 

31.  Give  a  corollary  of  this  theorem. 

32.  How  can  you  test  if  two  straight  lines  are  parallel? 


1  In  the  review  we  shall  always  (as  here)  simply  state  the  theorem  to  be  proved 
or  problem  to  be  solved  The  learner  is  to  understand  that  the  proof  (or  solution) 
is  required. 


72  GEOMETRY  FOR  BEGINNERS. 

83.    Two  angles  have  their  sides  respectively  parallel;   then, — 

(1)  If  these  sides  are  directed  the  same  way  from  the  vertex,  or  if  they 
are  directed  opposite  ways,  the  angles  are  equal. 

(2)  If  the  sides  are  directed,  the  one  pair  the  same  way,  the  other  pair 
opposite  ways,  the  sum  of  the  angles  is  180° 

EXERCISES. 

1.  What  is  the  ratio  of  one  right  angle  to  three  right  angles?  of  one-third 
of  a  right  angle  to  four  right  angles? 

2.  How  can  you  test  whether  two  intersecting  lines  are  perpendicular  to 
each  other? 

3.  How  can  you  test  whether  the  two  edges  of  a  square  are  truly  at  right 
angles  to  each  other? 

Suggestion.  —  How  was  the  straight  edge  of  a  ruler  tested  (§  23,  Exercise  6)? 

4.  Make  any  angle,  and  then,  by  the  aid  of  a  protractor,  make  an  angle 
four  times  as  large  ;   also,  an  angle  one-fourth  as  large. 

5.  Mark  three  points,  join  them  by  lines,  then  measure  the  three  angles 
which  the  lines  make  with  one  another.     What  is  their  sum? 

6.  Can  you  make  three  angles  with  two  lines? 

7.  If  two  lines  intersect,  and  one  of  the  four  angles  is  a  right  angle,  prove 
that  the  other  three  are  also  right  angles. 

8.  Draw  five  lines  meeting  at  a  point ;    how  many  angles  are  formed?  what 
is  their  sum?     If  the  angles  are  equal,  find  the  value  of  each  in  degrees. 

9.  Ten  lines  meet  at  a  point  so  as  to  form  a  regular  ten-rayed  star ;   find  the 
value  of  the  angle  between  any  two  rays. 

10.  What  is  the  supplement  of  i°?  of  179°?  of  180°? 

11.  Of  two  adjacent  angles  the  greater  is  twice  the  less  ;   find  the  values  of 
the  angles  in  degrees.     What  is  the  ratio  of  each  to  four  right  angles? 

12.  The  bisectors  of  adjacent  angles  are  at  right  angles  to  each  other. 
18.    The  bisector  of  one  of  two  vertical  angles  bisects  the  other  also. 

14.  If  two  parallel  lines  are  cut  by  a  third  line,  then  (a)  the  bisectors  of  two 
alternate  angles  are  parallel  to  each  other;  (U}  the  bisectors  of  two  op- 
posite angles  are  perpendicular  to  each  other. 

15.  What  is  the  difference  between  a  theorem  and  an  axiom?   between  a 
theorem  and  a  corollary  ? 


§59-] 


CHAPTER  iv. — TRIANGLES. 


73 


CHAPTER  IV. 
TRIANGLES. 

CONTENTS.  — I.  Sides  of  a  Triangle  (§§  59-63)-  II.  Angles  of  a  Triangle  ($$ 
64-66).  III.  Similarity,  Equivalence,  and  Equality  (§  67,  68).  IV.  Equal  Tri- 
angles (§$  69-79).  V.  Some  Consequences  of  the  Equality  of  Triangles  (§§ 
80-93).  VI.  Applications  (§§  94-98). 

L—  Sides  of  a  Triangle. 

§  59.  In  Fig.  56  are  several  figures  enclosed  on  all  sides  by 
lines.  Each  is  a  portion  of  a  plane  surface  (the  surface  of  the 
paper) ,  and  is  called  a  Plane  Figure. 


Fig.  56. 

Definition  I.  — A  PLANE  FIGURE  is  a  portion  of  a  plane  sur- 
face bounded  on  all  sides  by  lines. 

Many  surfaces  with  which  we  are  familiar,  —  for  example,  the  out- 
side of  our  houses,  the  floor  and  ceiling  of  a  room,  the  leaf  of  a 
book,  and,  in  many  cases,  gardens,  fields,  parks,  etc.,  —  furnish  in- 


74  GEOMETRY    FOR   BEGINNERS.  [§  60. 

stances  of  plane  figures  bounded  by  straight  lines.  Such  figures 
are  called  Polygons. 

Definition  II. — A  POLYGON  is  a  plane  figure  bounded  by 
straight  lines. 

The  bounding  lines  are  called  the  SIDES  of  the  polygon,  and 
their  sum  is  called  the  PERIMETER  of  the  polygon. 

Exercises.  —  1.  In  Fig,  56  which  figures  are  polygons  and  which  are 
not  polygons? 

2.  How  many  sides  and  how  many  angles  has  each  polygon  in  Fig.  56  ? 

3.  Draw  a  polygon  of  five  sides  ;   six  sides  ;   eight  sides  ;   ten  sides.     How 
many  angles  has  each? 

§  60.  Polygons  receive  different  names  according  to  the  num- 
ber of  their  sides. 

At  least  three  straight  lines  are  required  to  enclose  completely  a 
part  of  a  plane  surface  :  the  two  sides  of  an  angle  are  not  sufficient. 
Therefore  a  polygon  cannot  have  less  than  three  sides. 

Definition. — A  polygon  of  three  sides  is  called  a  TRIANGLE 

(Fig- 57)- 

A  triangle  is  usually  denoted  by  three  letters  standing  at  the 
vertices  of  its  angles.  Instead  of  writing  the  word  triangle,  the 
sign  A  is  sometimes  employed.  Thus  the  triangle  in  Fig.  57  is 
called  the  triangle  A  B  C,  or  A  A  B  C. 

Every  triangle  has  six  PARTS,  three 
sides,  AB,  AC,  B  C  (Fig.  57),  and 
three  angles,  A,  B,  and  C. 

Each  side,  as  A  B,  has  two  adja- 
cent  angles,  A  and  B,  and  one  oppo- 
site angle,  C ;  each  angle,  as  A,  is  included  between  two  sides, 
A  B  and  A  C,  and  is  opposite  to  the  third  side,  B  C.* 

Exercises.  —  1.  What  angles  are  adjacent  to  the  side  A  C  (Fig.  57)  ?  to 
the  side  B  C ?  What  angle  is  opposite  to  B  C? 

2.  By  what  sides  is  the  angle  B  included?  the  angle  C?  What  side  is 
opposite  to  B?  to  C? 


§6i.] 


CHAPTER   IV. -^TRIANGLES. 


§  61.  If  we  prolong  one  side  of  a  triangle,  the  prolongation 
makes  with  the  adjoining  side  an  an- 
gle called  an  EXTERIOR  ANGLE  of  the 
triangle,  and  the  three  angles  of  the 
triangle  are  termed,  by  way  of  dis- 
tinction, INTERIOR  ANGLES. 

In  Pig.  j8  m  is  an  exterior  angle,  Fi£*  58- 

b  is  the  adjacent  interior  angle ;  a  and  c  are  the  opposite  interior 
angles. 

Exercises.  — 1.  Draw  a  triangle,  and  then  prolong  each  side  in  both  direc- 
tions. How  many  exterior  angles  are  formed?  For  each  exterior  angle  point 
out  the  adjacent  interior  angle,  and  also  the  opposite  interior  angles. 

2.    If  an  exterior  angle  of  a  triangle  is  75°,  find  the  adjacent  interior  angle. 

§  62.  Theorem.  —  The  sum  of  two  sides  of  a  triangle  is  al- 
ways greater  than  the  third  side. 

For  the  straight  line  AB  (Pig.  57)  is  the  shortest  line  from 
AtoB  (§  32),  therefore  less  than  the  broken  line  A  C  B  ;  that  is, 
A  C  +  CB>  AB.  For  the  same  reason  it  follows  that  A  C  + 
AB>CB,  and  that  A  B  +  B  C>  A  C. 

Triangles  are  divided  into  three  classes,  according  to  the  lengths 
of  their  sides  :  the  EQUILATERAL  (Pig.  59,  I.),  in  which  the  three 


III. 


Fig-  59- 


sides  are  equal,  the  ISOSCELES  (Pig.  59,  II.),  in  which  two  of  the 
sides  are  equal,  and  the  SCALENE  (Pig.  59,  IH.)>  in  which  the  sides 
are  all  unequal. 


GEOMETRY  FOR  BEGINNERS. 


[§63. 


Exercises.  —  1.    Can  a  triangle  have  for  its  sides.  2m,  3™,  and  6m?    8m,  4™, 
and  2m?     im,  2m,  and  3™?    3™,  5™,  and  ym? 

2.  If  the  perimeter  of  a  triangle  is  2m,  what  is  the  greatest  value  which  one 
side  can  have? 

3.  Draw,  free-hand  (#),  an  equilateral;    (£),  an  isosceles  ;    (c),  a  scalene 
triangle. 

4.  The  perimeter  of  an  equilateral  triangle  is  1.5™;   find  each  side. 

5»    Can  you  draw  accurately  with  the  dividers  an  equilateral  triangle  of 
given  side,  say  6ocm? 


§  63.  We  may  regard  a  triangle  as  resting  on  one  of  its  sides 
as  a  base ;  this  side  is  then  called  the  BASE  of  the  triangle  ;  the 
vertex  of  the  angle  opposite  to  the  base  is  called  the  VERTEX  of 
the  triangle  ;  and  the  perpendicular  line  drawn  from  the  vertex  to 
the  base  (or  base  prolonged)  is  called  the  ALTITUDE  of  the  tri- 
angle. If  the  triangle  has  a  horizontal  side,  this  is  usually  taken 
for  the  base. 

Thus  in  the  triangle  ABC  (Fig.  60)  we  would  naturally  take 

the  side  A  B  for  the  base ; 
then  C  is  the  vertex,  and  CD 
is  the  altitude.  If  A  C  is 
taken  as  the  base,  B  is  the 
vertex,  and  B  E  the  altitude. 
If  B  C  is  taken  as  the  base, 
A  is  the  vertex,  and  A  F  the 
altitude. 


D 


N 


M 
Fig.  60. 

Notice  that  the  three  altitudes  intersect  in  one  point. 

The  altitude  may  be  outside  the  triangle ;  this  is  the  case  in 
the  triangle  MNO,  in  which  MJVis  the  base,  and  OP  the  altitude. 
The  altitude  must  fall  outside  if  the  triangle  has  an  obtuse  angle, 
and  one  of  the  sides  of  this  angle  is  taken  as  the  base. 

In  an  isosceles  triangle,  the  side  unequal  to  the  other  sides  is 
always  taken  as  the  base,  and  the  equal  sides  are  usually  referred 
to  as  the  sides  of  the  triangle. 


64.] 


CHAPTER   IV.  —  TRIANGLES. 


77 


Exercises.  —  1.  Draw  a  triangle  having  all  its  angles  acute,  and  then 
draw  the  three  altitudes. 

2.  Draw  a  triangle  having  an  obtuse  angle,  and  then  draw  the  three  altiv 
tudes.     How  many  of  the  altitudes  lie  outside  the  triangle? 

3.  Draw  an  isosceles  triangle  and  its  altitude. 

4.  On  a  given  line  as  base,  how  many  isosceles  triangles  can  be  con. 
structed?     How  many  equilateral  triangles? 

II.— Angles  of  a,  Triangle. 

§  64.  In  order  to  find  the  sum  of  the  angles  of  a  triangle  ABC 
(Fig.  61),  let  us  bring  them  all  to  a  common  vertex.  For  this 
purpose,  suppose  a  line  D  E  drawn  through  the  vertex  C  of  the  tri- 
angle parallel  to  the  base  A  B :  we  thereby  make  the  angles  ;;/ 
and  ;z.  The  angles  m  and  a  are  equal  because  they  are  alternate 
angles  (§  56),  and  the  angles  n  and  b  are  equal  for  the  same 


Fig.  61. 


reason.  Hence,  a  +  b-\-c=m-\-n  +  c  (Axiom  II.).  But 
m  -f-  n  -h  c  =  1 80°,  or  a  straight  angle.  Therefore,  a  +  b  +  c  = 
1 80°  (Axiom  I.). 

This  proof  applies  equally  well  to  any  triangle  ;  therefore,  to  all 
triangles  ;  hence  :  — 

Theorem.  —  The  sum  of  the  angles  of  a  triangle  is  eqital  to 
two  right  angles,  or  180°. 

Illustration.  —  The  truth  of  this  theorem  may  be  shown  to 
the  eye  as  follows  :  Cut  out  of  paper  a  triangle,  ABC  (Fig.  62), 
and  draw  its  altitude  CD.  Then  fold  over  the  corners  on  the 


78  GEOMETRY   FOR   BEGINNERS.  [§65. 

dotted  lines  as  edges.  This  will  bring  the  vertices  of  the  three 
angles  to  the  same  point  D,  and  we  shall  see  that  the  three  angles 
together  just  make  a  straight  angle. 

In  the  same  way  we  might  test  one  triangle  after  another,  find- 
ing in  each  case  that  the  sum  of  the  angles  was  180°  ;  after  a  time 
we  should  become  convinced  that  all  triangles  were  subject  to  this 
law. 

But  in  the  proof  given  above  a  single  case  is  sufficient ;  for  a 
little  reflection  shows  that  the  same  reasoning  holds  good  for  all 
cases. 

§  65.  From  the  preceding  important  theorem  several  corolla- 
ries follow. 

Corollaries.  —  I.  The  sum  of  two  angles  of  a  triangle  must 
always  be  less  than  180°. 

2.    A  triangle  can  have  only  one  right,  or  one  obtuse,  angle. 

In  other  words,  two  angles  of  a  triangle  must  be  acute ;  the 
third  may  be  either  acute,  right,  or  obtuse.  If  it  is  acute,  the  tri- 
angle is  called  an  ACUTE  TRIANGLE  (Fig.  6j,  I.)  ;  if  it  is  right,  the 


I.  II. 

Fig.  63. 

triangle  is  called  a  RIGHT  TRIANGLE  (Fig.  63,  II.)  ;  if  it  is  obtuse, 
the  triangle  is  called  an  OBTUSE  TRIANGLE  (Fig.  63,  III.). 

In  a  right  triangle  the  side  opposite  to  the  right  angle  is  called 
the  HYPOTENUSE,  and  the  other  sides  are  called  the  LEGS. 

3.  If  two  angles  of  a  triangle  are  known,  the  third  angle  will 
be  found  by  subtracting  the  sum  of  the  two  known  angles  from 
1 80°. 


§  66.]  CHAPTER   IV.  —  TRIANGLES.  79 

4.  If  two  angles  of  one  triangle  are  equal  respectively  to  two 
angles  of  another  triangle,  then  the  third  angle  of  the  one  is  equal 
to  the  third  angle  of  the  other. 

5.  In  a  right  triangle  the  sum  of  the  acute  angles  is  equal  to  a 
right  angle  or  90°. 

Hence,  if  one  of  the  acute  angles  in  a  right  triangle  is  known, 
the  other  can  be  found  by  subtracting  the  first  from  90°. 

Definition. — If  the  sum  of  two  angles  is  90°,  each  is  called 
the  COMPLEMENT  of  the  other. 

The  acute  angles  of  a  right  triangle  are  always  complements  of 
each  other. 

Exercises.  — 1.   Can  a  triangle  have  one  right  aw^/one  obtuse  angle? 

2.  If  one  leg  of  a  right  triangle  is  taken  as  base,  what  is  the  altitude? 

3.  Two  angles  of  a  triangle  are,  — 

O)  37°  and  71°;  (V)    45°  32'  18"  and  62°  50'  57"; 

(£)   82°  and  48°;  O)    64°  47'  33"  and  77°  18'  41"; 

(<)   40°  28'  and  18°  57';  (/)  179°    o'  54"  and    o°  59'    5"; 

find  the  third  angle  in  each  case. 

4.  If  one  acute  angle  of  a  right  triangle  is  (a)    30°;    (£)   45°;    (c]   63°; 
(</)   27°  15';    0)    58°  12'  48",  find  in  each  case  the  other  acute  angle. 

5.  What  is  the  complement  of  30°?  45°?  60° ?  89°?  90? 

§  66.  If  we  subtract  the  angle  b  (Fig.  64)  from  180°,  the  re- 
mainder is  equal  to  a  +  c  (§  64). 
But  this  remainder  is  also  equal  to  the 
exterior  angle,  m;  why  (§  53)? 
Therefore,  m  =  a  -f-  c  (Axiom  I.). 
That  is,  — 

Theorem.  —  An  exterior  angle  of  Fig- &*• 

a  triangle  is  equal  to  the  sum  of  the  two  opposite  interior  angles. 

Exercises.  —  1.  Find  the  exterior  angle  of  a  triangle  if  the  opposite  inte- 
rior angles  are  38°  35'  and  69°  46'. 

2.  An  exterior  angle  of  a  triangle  is  60°,  and  one  of  the  opposite  interior 
angles  is  30°;   find  the  other. 

3.  An  exterior  angle  of  a  triangle  is  90°.     What  kind  of  a  triangle  is  it? 

4.  Prove  that  the  sum  of  the  six  exterior  angles  of  a  triangle  is  720°. 


80  GEOMETRY   FOR   BEGINNERS.  [§  67. 

III.  —  Similarity,  Equivalence,  and  Equality. 

§  67.  We  may  compare  two  triangles  with  respect  either  to  their 
shape  (form)  or  to  their  size  {magnitude}. 

There  are  four  cases.     Either  (a) ,  the  triangles  have  the  same 
shape;    or  (£),  they  have  the  same  size;    or  (c),  they  agree  in 
both  shape  and  size  ;  or  (d)  ,  they  differ  in  both  shape»and  size. 
For  example  :  in  Fig.  65  the  triangles  ABC  and  P O  Q  have 
the   same   shape ;    ABC  and  D  E  F 
have  the  same  size  ;  A  B  C  and  M ' N  O 
agree  in  both  shape  and  size;  DEF 
and   P  O  Q  differ  in  both  shape  and 
size. 

If  two  triangles  have  the  same  shape, 
they  are  called  SIMILAR  TRIANGLES  ;  if 
they  have  the  same  size,  they  are  called 
EQUIVALENT  TRIANGLES  ;  if  they  agree 
both  in  shape  and  size  —  that  is,  if  they 
are  both  similar  and  equivalent  —  they  are  called  EQUAL  TRI- 
ANGLES. 

Exercises.  —  1.  To  which  class  of  triangles  do  P  O  Q  and  M  0  N  be- 
long? M  0  N  and  D  E  F? 

2.    Draw  (free-hand)  two  triangles  of  each  kind. 

§  68.  What  has  just  been  said  about  triangles  may  be  extended 
to  space  magnitudes  in  general. 

Two  straight  lines  always  have  the  same  shape  whatever  be  their 
lengths ;  so  have  two  circles,  two  cubes,  or  two  spheres,  however 
much  they  differ  in  size. v 

Again,  a  curved  line  may  have  the  same  length  as  a  straight 
line  ;  a  field  bounded  by  curved  lines  may  enclose  the  same  extent 
of  surface  as  a  field  bounded  by  straight  lines ;  a  cubical  vessel 
may  hold  the  same  quantity  of  water  as  a  vessel  shaped  like  a  cyl- 


§  69.]  CHAPTER   IV.  —  TRIANGLES.  81 

cylinder  or  a  sphere.     In  all  these  cases  the  size  is  the  same,  the 
shape  different. 

Lastly,  two  magnitudes,  whether  lines,  surfaces,  or  solids,  may 
have  at  once  the  same  size  and  the  same  shape. 

Definitions.  —  I.  Two  magnitudes  which  have  the  same  shape 
are  called  SIMILAR  MAGNITUDES. 

II.  Two    magnitudes   which   have   the  same   size    are    called 
EQUIVALENT  MAGNITUDES. 

III.  Two  magnitudes  which  agree  both  in  shape  and  in  size  are 
called  EQUAL  MAGNITUDES. 

Between  two  similar  magnitudes  we  place  the  sign  ^  ;  between 
two  equivalent  magnitudes  the  sign  =  ;  between  two  equal  mag- 
nitudes both  signs  ^  ;  or,  if  the  shape  is  not  taken  into  account, 
simply  the  sign  =. 

Of  these  three  kinds  of  agreement  between  space  magnitudes, 
that  of  equality  is  the  simplest,  and  ought  therefore  to  be  studied 
first. 

IV.—  Equal  Triangles. 

§  69.  Since  two  equal  triangles  agree  both  in  size  and  in  shape, 
they  can  differ  from  each  other  only  in  their  position  in  space,  and 
must,  when  one  is  laid  on  the  other,  coincide  in  all  their  six  parts  ; 
in  other  words,  their  three  sides  and  three  angles  must  be  equal 
each  to  each. 

Compare,  for  example,  the  equal  triangles  ABC  and  M  '  N  O 


Of  the  sides,  A  B  =  MN;  A  C  =  MO;  B  C  =  N  O. 

Of  the  angles,  C  =  O;  B  =  N;  A  =  M. 

In  two  equal  triangles  eqtial  sides  are  opposite  to  equal  angles, 
and  equal  angles  are  opposite  to  equal  sides. 

In  any  two  triangles  by  corresponding  sides  (angles)  are  meant 
those  which  are  opposite  to  equal  angles  (sides). 


82  GEOMETRY    FOR    BEGINNERS.  [§    'JO. 

§  70.  There  would  be  nothing  more  to  say  about  equal  triangles 
were  it  not  for  the  fact  that,  when  in  two  triangles  a  certain  number 
of  the  six  parts  are  equal  each  to  each,  the  remaining  parts  must 
also  be  equal,  and  the  triangles  must  be  equal.  This  fact  is  one 
of  the  corner-stones  in  the  science  of  Geometry  ;  upon  it  are  based 
the  proofs  of  many  important  theorems,  and  it  is  likewise  a  fact 
of  great  practical  value  in  the  various  uses  to  which  Geometry  can 
be  applied.  (See  Part  VI.  of  the  present  chapter.) 

Let  us  therefore  proceed  to  find,  once  for  all,  how  many  and 
what  parts  in  two  triangles  must  be  equal  in  order  to  make  the  tri- 
angles equal ;  in  other  words,  how  many  and  what  parts  of  a  tri- 
angle must  be  given  in  order  to  determine  the  size  and  shape  of 
the  triangle. 

1.  If  only  one  part,  a  side  or  an  angle,  is  given,  we  can  con- 
struct as  many  different  triangles  as  we  please,  all  having  this  part. 
Therefore,  by  one  part  the  size  and  s-hape  of  a  triangle  is  not 
determined. 

2.  Likewise,  if  two  parts  are  given,  whether  two  sides,  two  angles, 
a  side  and  an  adjacent  angle,  or  a  side  and  the  opposite  angle,  it 
is  easy  to  show  that  an  indefinite  number  of  triangles  may  be 
made,  all  having  these  two  parts  and  yet  differing  in  their  other 
parts  (see  Exercises  3-6  below) .     Hence  two  parts  are  not  suffi- 
cient to  determine  the  size  and  shape  of  a  triangle. 

3.  If  three  parts  are  given,  they  may  be,  — 
(/.)     The  three  angles  ; 

(«'.)    One  side  and  two  angles  ; 

(tit.)  Two  sides  and  the  included  angle ; 

(«/.)   Two  sides  and  an  angle  opposite  to  one  of  them ; 

(v.)    The  three  sides. 

These  cases  require,  each  of  them,  a  special  investigation. 

Exercises.  —  1.    Draw  three  triangles  having  a  common  side. 

2*    Draw  three  triangles  having  a  common  angle. 

3.    Draw  two  triangles,  both  having  for  two  of  their  sides  4Ocm  and  70°™. 


'§  7'.] 


CHAPTER    IV. TRIANGLES. 


83 


4.  Draw  two  triangles,  both  having  for  one  side  50°™,  and  for  an  adjacent 
angle  30°. 

5.  Draw  three  triangles,  all  having  for  one  side  45cm,  and  for  the  opposite 
angle  60°. 

Solution.  —  Make  an  angle  of  60°.  Then  take  45°™  between  the  dividers,  place 
one  point  on  one  side  of  the  angle,  the  other  point  on  the  other  side.  Join  these 
two  points  and  the  triangle  is  constructed.  The  other  triangles  are  made  in  the 
same  way,  taking  care  to  use  different  points  on  the  sides  of  the  angle. 

6.  Draw  two  triangles,  both  having  the  angles  30°  and  60°.     What  kind 
of  triangles  are  they  ? 


§  71.  Is  a  triangle  determined  if  we  know  its  three  angles? 
Compare  the  triangles  ABC  and  DEF 
(Fig.  66}  ;  as  regards  their  angles, 
A  =  D,B  =  E,  C  =  F  (§  58)  ;  that 
is,  the  triangles  have  their  angles  equal 
each  to  each.  Now  the  triangles  them- 
selves are  not  equal  ;  they  have  the 
same  shape,  but  they  differ  in  size.  A  B 

In  fact  we  may  make  as  many  tri-  F-    66 

angles  as  we  please,  all  having  the  same 
shape,  but  differing  in  size  ;  therefore,  — 

By  the  three  angles  a  triangle  is  not  determined. 

Two  triangles  which  have  their  angles  equal  <each  to  each  are 
said  to  be  mutually  equiangular. 

Exercises.  —  1.    Draw  two  mutually  equiangular  triangles. 

2,    Can  you  draw  two  mutually  equiangular  triangles  which  differ  in  shape? 


§  72.  Problem.  —  To  construct  a  triangle,  having  given  a  side 
and  two  angles. 

When  two  angles  are  given,  the  third  can  always  be  found 
(how?).  We  shall  therefore  suppose  that  the  given  angles  are 
both  adjacent  to  the  side. 

Let  a  (Fig.  6?)  be  the  given  side,  and  60°  and  40°  the  given 


84  GEOMETRY    FOR    BEGINNERS.  [§    72« 

angles.     Draw  a  line  A  B  —  a.     At  the  points  A  and  B  make 
angles  equal  respectively  to   60°  and  40°.     The  sides  of  these 
angles  will  meet  at  a  point,  C,  and  A  B  C  is  the  triangle  required. 
Why  cannot  these  sides  be  parallel,  and  hence  never  meet? 


fif.67.      a          M 

If  we  now  construct  another  triangle,  MN  O,  having  the  same 
three  parts,  it  will  be  equal  to  the  triangle  ABC. 

Proof.  —  Place  MN  O  on  A  B  C,  so  that  MN  falls  wAB.  M 
will  fall  on  A,  and  N  on  Bt  since  MN  =  A  B  =  a.  The  lines 
MO  and  A  C  will  coincide  in  direction,  since  the  angles  at  J/and 
A  are  equal  (each  60°),  and  the  lines  N  O  and  B  Cwill  also  coin- 
cide in  direction,  since  the  angles  at  N  and  B  are  equal  (each  40°). 
Therefore  O  must  coincide  with  C,  and  the  triangles  will  coincide 
in  all  their  parts.  Point  out  the  parts  which  are  equal  each  to  each. 

The  construction  and  proof  would  be  the  same  for  other  values 
of  the  given  sides^and  angles  ;  therefore,  by  a  side  and  two  angles, 
a  triangle  is  completely  determined ;  and  also,  — 

Theorem  (I.  Law  of  Equality).  — If  in  two  triangles  a  side 
and  two  angles  are  equal  each  to  each,  the  triangles  are  equal. 

Remark.  — The  method  used  above  to  prove  the  two  triangles 
equal  is  called  the  Method  of  Superposition.  When  it  can  be  em- 
ployed to  prove  the  equality  of  two  magnitudes,  it  is  a  very  easy 
method,  if  we  only  take  care  to  bear  in  mind  what  parts  in  the 
two  magnitudes  are  known  to  be  equal. 

Exercises.  —  1.  Draw  to  reduced  scale  (on  paper,  I  :  2000;  on  the  black- 
board, 1 :  200)  a  triangle,  one  side  of  which  is  220'",  and  the  adjacent  angles 
70°  and  50°. 


§    73-]  CHAPTER   IV. TRIANGLES.  85 

2.  Draw  to  reduced  scale  the  outlines  of  a  triangular  field  of  which  one 
side  is  6oom,  and  the  adjacent  angles  80°  and  47°. 

NOTE. — Always  write  the  scale  employed  by  the  side  of  the  figure. 

3o  The  same  Exercise,  one  side  of  the  field  being  40™,  an  adjacent  angle 
being  65°,  and  the  opposite  angle  being  37°. 

4.  Draw  a  right  triangle  having  given,  — 

(2.)      One  leg  =  8dm,  and  the  adjacent  acute  angle  =  57°. 

(?'/.)    One  leg  =  6ocm,  and  the  opposite  angle  =  25°. 

(z'zY.)  The  hypotenuse  =  90°™,  and  an  adjacent  angle  =  42°. 

5.  What  limit  is  there  to  the  sum  of  the  two  given  angles  in  the  problem 
of  this  section  in  order  that  the  solution  may  be  possible  ? 

6.  What  are  the  given  things  or  data 1  in  the  problem  of  this  section  ? 

§  73.  Problem.  —  To  construct  a  triangle  having  given  two 
sides  and  the  included  angle. 

Let  a  and  b  (Fig.  68}  be  the  given  sides,  and  50°  the  included  angle. 

Make  an  angle  A  =  50°,  take  A  B  =  a,  A  C  =  b,  and  join  BC. 
A  B  C  is  the  triangle  required. 
Draw  another  triangle  having 
the  three  given  parts,  and  then 
prove  by  superposition  that  it 
must  be  equal  to  A  B  C. 

Hence,  by  two  sides  and  the 
included  angle  a  triangle  is  com- 
pletely determined ;  and,  — 

Theorem  (II.  Law  of  Equality).  — If  in  two  triangles  two 
sides  and  the  included  angle  are  equal  each  to  each,  the  triangles 
are  equal. 

Exercises.  — 1.  Draw  (to  reduced  scale)  a  triangle  with  the  sides  49™  and 
77m,  and  the  included  angle  45°. 

2.  Represent  (to  scale)  on  paper,  and  also  on  the  blackboard,  a  triangle, 
two  sides  of  which  are  140™  and  ioom,  and  the  included  angle  55°.  Are  the 
two  figures  equal,  equivalent,  or  similar?  Are  they  mutually  equiangular? 

1  The  singular  of  data  is  datum.  The  word  is  from  the  Latin,  and  means 
given. 


86  GEOMETRY   FOR   BEGINNERS.  [§   74. 

3.  In  an  isosceles  triangle  one  of  the  equal  sides  is  32cm,  and  the  angL 
at  the  vertex  is  82°;   draw  the  triangle. 

4.  Draw  a  right   triangle  whose  legs  are   i6cm  and  2Ocm.     What  three 
parts  are  here  known? 

5.  Draw  an  isosceles  right  triangle  whose  legs  are  i6cm  each.     What  are 
the  three  parts  here  given? 

6.  What  is  the  greatest  value  which  the  given  angle  in  the  problem  of 
this  section  can  have? 

7»    What  are  the  data  in  the  problem  of  this  section? 

§  74.  Problem.  —  To  construct  a  triangle  having  given  two 
sides  and  an  angle  opposite  to  one  of  them. 

There  are  two  cases  according  as  the  given  angle  is  opposite 
(/'.)  to  the  greater,  or  (it.)  to  the  less  side. 


Fig.  69. 

Case  (/.).  Let  a  and  b  (Fig.  69)  be  the  given  sides,  and  let 
a  >  b ;  also  let  the  angle  opposite  to  a  be  70°.  Make  an  angle 
CAB  =  70°,  and  lay  off  A  C  —  b  \  this  determines  two  corners, 
A  and  C,  of  the  triangle.  The  third  corner  must  lie  in  the  line 
A  B,  and  its  distance  from  C  must  be  equal  to  a ;  therefore,  it 
must  also  lie  in  a  circumference  described  from  C  as  centre  with 
a  radius  equal  to  a.  Hence  it  must  be  the  intersection  of  this 
circumference  with  the  line  AB.  Now  this  circumference  cuts 
A  B  in  two  points,  B  and  B'9  so  that  we  obtain  two  triangles, 
A  B  C  and  A  B'  C.  Of  these,  however,  only  A  B  C  has  the  three 
given  parts;  A  B' C  has,  it  is  true,  the  given  sides,  but  not  the 
given  angle  ;  therefore,  it  is  not  the  triangle  required. 


8   74']  CHAPTER   IV. TRIANGLES.  87 

If  we  construct  another  triangle,  M ' N  O,  having  the  same  three 
jarts,  it  will  be  equal  to  the  triangle  ABC. 

Proof.  —  Place  A  MN  O  on  A  A  B  C,  making  the  equal  sides 
M  O  and  A  C  coincide.  MN  will  coincide  in  direction  with  A  B 
because  OMN=  CAB  =  70°  ;  therefore  N  will  fall  in  AB.  But 
N  must  also  lie  in  a  circumference  having  (now)  C  for  centre,  and 
radius  =  a  ;  therefore  N  will  coincide  with  B.  And  ON  will 
coincide  with  CB,  and  the  triangles  will  coincide  in  all  their  parts. 
Hence  we  see  that  by  two  sides  and  the  angle  opposite  to  the 
greater  side  a  triangle  is  completely  determined ;  and  that,  — 

Theorem  (III.  Law  of  Equality).  —  If  in  two  triangles 
two  sides  and  the  angle  opposite  to  the  greater  side  are  equal  each 
to  each,  the  triangles  are  equal. 

Case  (//.).  Let  a  and  b  (Fig.  70)  be  the  given  sides,  and  let 
a  <  b ;  also  let  the  angle  op- 
posite to  a  =  42°. 

By  proceeding  as  in  Case 
(/.)  we  obtain  two  triangles, 
AB  C  and  A  B'C,  both  hav- 
ing the  three  given  parts,  yet 

differing  in  size  and  in  shape.  A  £'\ .-~''B 

Therefore,  by  two  sides   and  Fi£-  7°> 

the  angle  opposite  to  the  less  side,  a  triangle  is  not  determined. 

There  are,  however,  only  two  solutions,  the  triangles  ABC  and 
AB'C. 

.  Exercises.  —  1.   In  the  triangles  A  BC  and  AJB'C  {Fig.  69)  point  out  the 
equal  parts  ;   also  the  unequal  parts. 

2.  What  relation  exists  between  the  angles  B  AC  and  &A  C  (Fig.  69)  ? 

3.  What  kind  of  triangles  are  the  triangles  BC '£'  and  N  ON1  (Fig.  69)  ? 

4.  What  equal  parts  have  the  triangles  AC  B  and  ACB1  (Fig.  70)  ? 

5.  Draw  a  triangle  with  the  sides  6ocm  and  9Ocm,  and  the  angle  opposite 
the  greater  side  76°. 

6.  Construct  a  right  triangle  the  hypotenuse  of  which  is  8dm,  and  one  leg 
is  6^™.     What  are  the  three  given  parts  ? 


GEOMETRY    FOR    BEGINNERS. 


[§75- 


7.  Construct  a  triangle  with  the  sides  8dm  and  4dm,  and  the  angle  opposite 
to  the  greater  side  80°. 

8.  When  are  the  data  in  Case  («'.)  of  this  section  such  that  no  triangle 
is  possible? 

Ans.  —  When  the  less  side  is  shorter  than  the  perpendicular  from  C  to  A  B 
(Fig.  70) ;  also  when  the  given  angle  is  either  right  or  obtuse. 

9.  When  the  less  side  is  just  equal  to  the  perpendicular  from  C  to  A  B, 
what  kind  of  a  triangle  is  obtained? 

§  75.  Problem.  —  To  construct  a  triangle  having  given  its 
three  sides. 

Let  a  b  c  (fig.  71)  be  the  given  sides.  Make  A  B  —  a  ;  then  A 
and  B  are  two  corners  of  the  triangle,  and  A  B  is  one  side.  Since 
the  third  corner  must  be  at 
the  distance  b  from  A,  it  must 
lie  in  a  circumference  described 
from  A  as  centre  with  the  ra- 
dius b.  For  a  like  reason  it 
must  also  lie  in  a  circumfer- 
ence described  from  B  as  cen- 
tre with  the  radius  c.1 

Therefore  it  must  be  at  the 
intersection  of  these  circum- 
ferences. Now  these  circum- 
ferences cut  each  other  in  two  points,  C  and  C',  one  above  the 
other  below  A  B ;  so  that  we  obtain  two  triangles,  ABC  and 
ABC',  both  having  the  three  given  sides. 

These  triangles,  however,  are  equal.  For,  if  we  fold  over  ABC' 
on  A  B,  C'  will  be  brought  above  A  B,  and,  as  it  remains  the  inter- 
section of  circumferences  having  A  and  B  for  centres,  and  b  and  c 
for  radii,  it  will  fall  on  C,  and  the  triangles  ABC  and  ABC1  will 
then  coincide.2 

1  It  is  obvious  that  in  solving  this  problem  entire  circumferences  are  not  re- 
quired ;  short  arcs  near  the  points  of  intersection  are  sufficient. 

2  See  Exercise  5,  page  92. 


Fig.  71. 


§    76.]  CHAPTER   IV. TRIANGLES.  89 

If  we  construct  another  triangle  with  the  same  three  sides,  it  will 
also  be  equal  to  ABC.  For  we  can  place  it  on  A  B  C  so  that 
one  of  its  sides  will  coincide  with  AB,  and  then  its  third  corner 
must  fall  on  C  for  the  same  reason  that  C'  falls  on  C  when  ABC* 
is  folded  over  on  A  B. 

From  what  precedes  we  conclude  that,  by  the  three  sides  a  tri- 
angle is  completely  determined ;  also  that,  — 

Theorem  (IV.  Law  of  Equality). — If  in  two  triangles  the 
three  sides  are  equal  each  to  each,  the  triangles  are  equal. 

Exercises.  —  1.  The  sides  of  a  triangular  garden  measure  8om,  65™,  and 
45m.  Draw  to  scale  a  plan  of  the  gai'den. 

2.  The  sides  of  a  triangular  field  are  2Oom,  240™,  and  300™.     Draw  to 
scale  a  plan  of  the  field. 

3.  Construct  a  triangle  whose  sides  are  2Ocm,  4Ocm,  and  6ocm. 

4.  Construct  a  triangle  whose  sides  are  2Ocm,  3Ocm,  and  6ocm. 

5.  When  is  the  solution  of  the  problem  of  this  section  impossible?     (See 
§62.) 

6.  Construct  an  isosceles  triangle  whose  base  is  7Ocm,  and  the  sum  of 
whose  other  sides  is  2m. 

7.  Construct  an  equilateral  triangle  one  side  of  which  is  im. 

8.  Construct  a  triangle  with  the  sides  3Ocm,  3Ocm,  and  3OCU1.     What  kind 
of  a  triangle  is  it? 

9.  Construct  a  triangle  with  the  sides  25cm,  25cm,  and  40°'".     What  kind 
of  a  triangle  is  it? 

10.  Construct  a  triangle  with  the  sides  2Ocm,  3Ocm,  and  4.ocm.     What  kind 
of  a  triangle  is  it? 

11.  Are  two  triangles  equal  if  they  have  equal  perimeters  ?    (Compare  the 
last  three  exercises.) 

§  76.    Problem.  —  To  make  a  triangle  equal  to  a  given  triangle. 

Take  any  three  parts  of  the  given  triangle  that  completely  de- 
termine a  triangle,  and  construct  with  them  a  new  triangle  ;  it  will 
be  equal  to  the  given  triangle.  The  best  parts  to  take  for  this  pur- 
pose are  the  three  sides.  Show  hew  to  construct  the  triangle  with 
them  (§  75). 


90 


GEOMETRY   FOR    BEGINNERS. 


[§  77. 


Exercises.  —  1.    In  what  four  ways  can  three  parts  be  chosen  with  which 
to  construct  a  triangle? 

2.    Draw  a  triangle;   then  construct  another  equal  to  it. 


Fig.  72. 


§  77.   Problem.  —  At  a  given  point  D  (Fig.  72)  of  a  given 
line  D  G,  to  make  an  angle  equal  to  a  given  angle  BAG. 

Take  on  the  sides  of  the  given  angle  A  M 
=  AN,  and  join  MN.  Then  construct  (by 
the  last  problem)  a  triangle  D  E  F  equal  to 
the  triangle  AMN.  The  angle  D,  corres- 
ponding to  the  angle  A,  will  be  equal  to  it. 

Exercises.  —  1.  What  kind  of  a  triangle  is^  M N~! 
Is  it  necessary  for  the  solution  of  the  problem  to 
make  A  MN  a.  triangle  of  this  kind? 

2.  Make  an  angle,  and  then  make  another  equal 
to  it. 

3.  Make  an  angle  equal   to  a  given  angle  with 
j                             the  protractor. 

4»    Make  an  angle  equal  to  a  given  angle  with  the  ruler  and  the  square. 
5.    Make  (with  ruler  and  compasses)  an  angle  of  60°;   an  angle  of  120°. 

§  78.   Problem.  —  Through  a  given  point  C  (Fig.  73)  to  draw 
a  line  parallel  to  a  given  line  A  B. 

Through  C  draw  any  straight  line  CD,  cutting  AB  in  D.    Then 

draw  (by  the  last  problem)  C  P  so 

that   the    angle    DCP  =  CDB. 

CP  is  parallel   to   AB    (§   57). 

Exercises.  —  1.   Draw  a  line,  and  then 
\  through  a  point  draw  another  line  par- 

\ allel  to  the  first. 

2.   The  same  exercise,  using  ruler  and 
square  instead  of  ruler  and  compasses. 
3.    Through  the  vertices  of  the  angles  of  a  triangle  draw  lines  parallel  to 
the  opposite  sides,  and  prolong  them  until  they  meet.     What  kind  of  a  figure 
is  thus  formed? 


/D 


73- 


§    79-]  CHAPTER   IV. TRIANGLES.  $1 

§  79.  A  PROBLEM  is  a  construction  to  be  made  according  to 
geometrical  laws. 

There  is  a  general  agreement  that  in  the  Science  of  Geometry 
no  instrument  shall  be  employed  in  making  a  construction  except 
ruler  and  compasses ;  if  the  problem  cannot  be  solved  by  these 
means  it  is  not  regarded  as  belonging  to  the  science.  In  the  Art 
of  Geometry  —  that  is,  in  the  applications  of  Geometry  to  prac- 
tical purposes,  such  as  Drawing  and  Land-measuring  —  various 
other  instruments  which  are  found  convenient  are  used,  as  divided 
rules,  protractors,  theodolites,  etc. 

Thus,  in  practice,  the  problem  of  §  77  is  usually  solved  with  the 
protractor,  or  with  the  ruler  and  the  square,  and  the  problem  of 
§  78  with  the  ruler  and  the  square. 

In  making  a  construction,  all  lines  which  are  merely  interme- 
diate steps  between  the  data  and  what  is  required  should  be  drawn 
dotted  instead  of  full  (see  Figs.  72  and  fj) .  Such  lines  are  called 
auxiliary  lines. 


V.  —  Some  Consequences  of  the  Equality  of  Triangles. 

§  80.  Let  A  B  C  (Fig.  74)  be  an  isosceles  triangle  having  A  C 
=  B  C,  and  let  CD  be  a  line 
bisecting  the  angle  C.  This  line 
divides  the  triangle  ABC  into  two 
triangles,  A  C  D  and  B  C  D.  Com- 
pare these  two  triangles  ;  they  have 
a  common  side  CD,  the  side  A  C 
=  BC  (why  ?),  and  the  angle  c=d 
(why?).  Therefore  the  triangles 
are  equal  (II.  Law  of  Equality), 
and  among  the  equal  parts  we  have  a  =  b ;  hence,  — 

Theorem.  —  /;/  an  isosceles  triangle  the  angles  opposite  to  the 
equal  sides  are  equal. 


92  GEOMETRY    FOR    BEGINNERS.  [§  80. 

Corollary.  —  In  an  equilateral  triangle  all  the  angles  are  equal  ; 
therefore  each  angle  =  1  80°  :  3  =  60°. 

Why  does  this  corollary  follow  from  the  theorem? 

Remark.  —  A  theorem  consists  of  three  parts  :  — 

(/.)  The  HYPOTHESIS  (or  hypotheses),  or  that  which  is  given,  or 
assumed  as  true,  at  the  start. 

(ii.)  The  CONCLUSION  (or  conclusions),  or  that  which  is  to  be 
proved. 

(Hi.)  The  PROOF. 

In  the  above  theorem  the  hypothesis  is  :  A  C  —  B  C  (Fig.  74)  , 
and  the  conclusion  is  :  a  —  b.  The  proof  consists  in  showing  that 
the  triangles  A  C  D  and  BCD  are  equal,  and  hence  inferring  that 
a  =  b. 

Exercises.  —  1.  In  an  isosceles  triangle  the  bisector  of  the  angle  at  the 
vertex  is  perpendicular  to  the  base,  and  bisects  the  base. 

Hints.—  Use  Fig.  74.  The  two  hypotheses  are:  A  C  =  B  C,  and  CD  bisects 
A  CB.  The  two  conclusions  are  :  C£>±  A  B,  and  CD  bisects  A  B.  Proof:  show 
that  &ACD^  /\BCD\  hence  ;w  =  «  =  90°  (why  ?),  and  AD  —  BD. 


2.  The  perpendicular  let  fall  from  the  vertex  of  an  isosceles  triangle  to 
the  base  bisects  the  base,  and  also  the  angle  at  the  vertex. 

Hints.—  Here  the  hypotheses  are  :  A  C  =  B  C,  and  CD  _L  A  B.  The  conclu- 
sions are  :  CD  bisects  A  B,  and  CD  bisects  A  C  B.  Prove  by  showing  that  A  A  CD 
^  A  B  CD. 

3.  The  line  joining  the  vertex  of  an  isosceles  triangle  to  the  middle  of  the 
base  is  perpendicular  to  the  base,  and  bisects  the  angle  at  the  vertex. 

Hints.  —  In  this  case  what  are  the  hypotheses  ?  the  conclusions  ?  Prove  by 
showing  (by  IV.  Law  of  Equality)  that  A  A  CD  ^l\B  CD. 

4.  What  three  conditions  does  the  line  CD  {Fig.  74)  fulfil? 

5*  Prove  (with  the  aid  of  the  theorem  of  this  section)  that  the  triangles 
A  B  C  and  A  B  C'  {Fig.  77,  p.  88)  are  equal. 

Hint.  —  By  drawing  C  C1  we  obtain  two  isosceles  triangles,  A  C  C'  and  B  C  C'. 

6.  Find  the  base  angles  of  an  isosceles  triangle  if  the  angle  at  the  vertex 
is  (*.)  27°  35',  (it.}  75°  18',  (Hi.}  124°  40'. 

7.  Find  the  angle  at  the  vertex  if  one  of  the  base  angles  is  (i.)  15°  I4;> 
(ii.}  48°  7',  (Hi.}  83°  4'. 

8.  In  an  isosceles  right  triangle  find  the  base  angles. 


§  3  1.]  CHAPTER   IV.  -  TRIANGLES.  93 

9,  Can  you  devise  a  way  of  finding  the  height  of  an  object  (tree,  tower, 
etc.)  by  means  of  an  isosceles  right  triangle? 

10«  Show  that  in  an  isosceles  triangle  the  exterior  angles  made  by  prolong- 
ing the  base  are  equal,  and  that  the  exterior  angles  at  the  vertex  are  also 
equal. 

11.  Find  the  three  angles  of  an  isosceles  triangle  if  the  exterior  angle  at 
the  vertex  is  146°  19'. 

12.  Find  the  three  angles  of  an  isosceles  triangle  if  the  exterior  angle  at 
the  base  is  120°  50'. 

13.  Show  that  the  exterior  angles  of  an  equilateral  triangle  are  all  equal, 
and  find  the  value  of  one  of  them. 

14.  Construct  an  isosceles  triangle  having  given,  — 

(a)  The  base  and  an  adjacent  angle  ; 
(<£)    The  base  and  the  opposite  angle  ; 
(<r)    The  side  and  an  angle  at  the  base  ; 
(d}  The  side  and  the  angle  at  the  vertex. 


§  81.  Let  us  now  transpose  the  hypothesis  and  the  conclusion 
of  the  theorem  in  §  80  ;  in  other  words,  let  us  suppose  that  in  a 
triangle  ABC  (Fig.  75)  the  angles  A  and  B  are  equal,  and 
inquire  if  the  sides  opposite  to  these  angles  must  also  be  equal. 

Let  CD  bisect  the  angle  C. 
Then  the  triangles  A  CD  and 
BCD  are  mutually  equiangular 
(why  ?)  .  They  also  have  the  side 
CD  common,  therefore  they  are 
equal  (I.  Law  of  Equality),  and 
A  C  =  B  C.  Hence,  — 

Theorem.  —  If  in  a  triangle 
two  angles  are  equal,  the  opposite 
sides  are  also  equal,  and  the  triangle  is  isosceles. 

Corollary.  —  If  the  three  angles  of  a  triangle  are  equal,  the 
triangle  is  equilateral. 

Why  does  this  corollary  follow  from  the  theorem  ? 

Remark.  —  If  we  transpose  the  hypothesis  and  the  conclusion 


94 


GEOMETRY    FOR   BEGINNERS. 


[§   82. 


of  a  theorem,  we  obtain  a  new  theorem  called  the  CONVERSE  of  the 
original  theorem.  The  theorem  of  this  section,  for  example,  is  the 
converse  of  the  theorem  of  the  last  section. 

The  converse  of  a  true  theorem  is  not  necessarily  true.  Take, 
for  instance,  the  theorem,  two  vertical  angles  are  always  equal ;  the 
converse,  two  equal  angles  are  always  vertical  angles,  is  not  true. 

Exercises.  —  1.  In  an  isosceles  triangle  the  bisectors  of  the  base  angles 
form  with  the  base  another  isosceles  triangle. 

2»  Prove  the  converse  theorem  (namely,  if  in  a  triangle  the  bisectors  of 
the  base  angles  form  with  the  base  an  isosceles  triangle,  then  the  triangle  is 
also  isosceles). 


§  82.   We  will  now  compare  in  a  triangle  two  unequal  sides  and 
the  opposite  angles. 

Let  ABC  {Fig.  76)  be  an  isosceles  triangle  having  A  B  =  A  C 
and  A  B  C  =  A  C  B.     Prolong  A  C  to  any  point  D,  and  join  B  D. 

Then  A  B  D  is  a  triangle  having  A  D 
>AB.  Now  compare  the  opposite 
angles.  Since  ABD  =  ABC  + 
C BD,  therefore  ABD  >  ABC. 
And  since  A DB  =  ACB-CBD 
(§  66),  therefore  ADB  <  A  CB, 
andhence<^f^C.  Therefore.^ Z> 
>  A  D  B.  Here  it  is  quite  clear  that 
the  two  conditions,  A  D  >  A  B,  and 
A  B  D  >  A  D  B,  must  always  be  ful- 
filled together ;  hence,  — 

Theorem.  I. —  Of  two  angles  of  a  triangle  the  greater  is  oppo- 
site to  the  greater  side. 

Theorem  II.  (Converse  of  Theorem  I.). —  Of  ftvo  sides  of  a 
triangle  the  greater  is  opposite  to  the  greater  angle. 

Corollary.  —  In  a  right  triangle  the  hypothenuse  is  the  greatest 
side. 


Fig.  76. 


CHAPTER    IV. TRIANGLES. 


95 


Exercises,  —  1.    From  which  theorem  does  the  corollary  follow?     Why 
does  it  follow? 

2.    What  similar  corollary  follows  for  an  obtuse  triangle? 


D          F 

Fig.  77, 


§  83.  Let  CD  JL  AB  (Fig.  77),  and  let  C  E,  C F,  C  G,  be 
any  other  lines  drawn  from  C  to  the 
line  A  B.  These  lines  are  the  hy- 
pothenuses  of  the  right  triangles 
CDF,  CDF,  CD  G,  respectively, 
and  therefore  are,  each  of  them, 
greater  than  CD  (§  82,  Corollary)  ; 
therefore,  — 

Theorem  I.  —  Of  all  lines  which     - 

A 

can  be  drawn  from  a  point  to  a 
straight  line  the  perpendicular  is  the 
shortest. 

Corollary.  —  Hence,  the  distance  from  a  point  to  a  straight  line 
is  the  length  of  the  perpendicular  let  fall  from  the  point  to  the  line. 

If  DE  =  DF,  then  A  CDE  ^  A  CDF  (why?),  therefore 
C  E  =  CF;  that  is, — 

Theorem  II.  —  Two  oblique  lines  equally  distant  from  the  foot 
of  the  perpendicular  are  equal. 

Again,  D  G  >  D  F.  Likewise  the  angles  CFD  and  CGD 
are  each  acute  (why?).  And  CFG,  being  the  supplement  of 
CFD,  must  be  obtuse.  Therefore,  in  the  triangle  CFG  we 
have  CFG  >  CGF;  hence  C  G  >  CF  (§  82,  Theorem  II.). 
That  is,— 

Theorem  III. —  Of  two  oblique  lines  unequally  distant  from 
the  foot  of  the  perpendicular,  the  more  remote  is  the  greater. 

Exercises.  —  1.  In  each  of  these  three  theorems  what  is  the  hypothesis? 
What  the  conclusion? 

2.    State  and  prove  the  converse  of  Theorem  II. 


96 


GEOMETRY    FOR    BEGINNERS. 


[§84. 


§  84.    Let  AB  C  and  ABD  (Fig.  78)  be  two  isosceles  trian- 

gles having  a  common  base  AB. 
Join  C  D.  The  triangles  A  CD  and 
B  CD  are  equal  (IV.  Law  of  Equali- 
ty). Let  the  triangle  A  CD  be 
folded  over  CD  until  it  coincides 
with  the  triangle  C  B  D  ;  then  A 
will  fall  on  B,  and  all  the  lines  and 
angles  which  fall  upon  each  other 
will  be  equal.  Hence  we  have,  (*'.) 
#  =  £  and  ;•  =  </;  (ii.)AE  =  BE; 
(tit-)  m  =  n,  and  therefore  CD  _L 
AB. 

Theorem.  —  If  two  isosceles  triangles  have  a  common  base, 
the  line  drawn  through  their  vertices  (i.)  bisects  the  angles  at  the 
vertices  ;  (ii.)  bisects  the  base  ;  (iii.)  is  perpendicular  to  the  base. 


Exercises.  —  1.  Prove  this  theorem,  when  (as  in 
Fig.  79}  the  vertices  of  the  two  isosceles  triangles 
are  both  on  the  same  side  of  the  base. 

2.  If  any  number  of  isosceles  triangles  have  a  com- 
mon base,  the  line  passing  through  the  vertices  of  two 
of  the  triangles  will  pass  through  the  vertices  of  all 
the  triangles. 

3.  If  (Fig.  79)  CAE=  64°  and  D  A  E  =  36°,  find 
all  the  other  angles  in  the  figure. 


79' 


§  85.   Problem.  —  To  bisect  a  given  angle  B  A  C  (Fig.  So) . 

Analysts. — The  theorem  of  §  84  suggests  a  mode  of  solving 
this  problem.  If  we  construct  an  isosceles  triangle  having  the 
given  angle  B  A  C  for  the  angle  at  the  vertex,  and  then  upon  its 
base  construct  any  other  isosceles  triangle,  the  line  which  joins 
the  vertices  of  the  two  triangles  must,  by  §  84  (i.) ,  be  the  bisector 
required, 


§86.] 


CHAPTER    IV.  —  TRIANGLES. 


97 


Construction.  —  With  A  as  centre  describe  an  arc  cutting  the 
sides  of  the  given  angle  in  M  and  N.     With  A 

M  and  N  as  centres,  and  equal  radii,  describe 
arcs  intersecting  at  a  point  D.  Join  A  D  : 
it  is  the  bisector  required. 

Exercises.  —  1.  Prove,  directly,  that  A  AMD 
^A  AND  (Fig.  80),  and  hence  that  B  A  D  = 
CAD. 

2.  Make  an  acute  angle,  and  then  bisect  it. 

3.  Make  an  obtuse  angle,  and  then  bisect  it. 

4.  Draw  a  triangle,  and  then  bisect  the  three 
angles.       In    how   many   points    do    the    bisectors 
meet  one  another? 

5*    Divide  an  angle  into  four,  and  also  into  eight,  equal  parts. 

6.    Make,  with  ruler  and  compasses,  an  angle  of  30°;   an  angle  of  45°. 


§  86.    Problem.  —  To  bisect  a  given  straight  line  A  B  (Fig.  81) . 

Analysis.  —  If  we  construct  upon  A  B  as  a  base  any  two  isos- 
celes triangles,  the  line  joining  their 
vertices  must,  by  §  84  (ii.),  bisect  ^V^ 

AB. 

Construction.  —  With  A  and  B  as 
centres  and  equal  radii,  describe  arcs 
intersecting  above  A  B  in  C,  and  be- 
low A  B  in  D.  Join  C  D  :  it  bisects 
A  B  in  E. 


Fig.  81. 


Exercises.  —  1.    Bisect   a   line,  first  freehand,  then  with  ruler  and  coin- 
passes. 

2.  Bisect  the  three  sides  of  a  triangle,  and  then  join  the  points  of  bisec- 
tion to  the  vertices  of  the  opposite  angles.     In  how  many  points  do  the  join- 
ing lines  (called  the  medians  of  the  triangle)  intersect? 

3.  Divide  a  line  into  four,  and  also  into  eight,  equal  parts. 

4.  Can  you,  by  means  of  this  problem,  divide  a  line  into  six  equal  parts? 

5.  Bisect  a  line  more  than  twice  as  long  as  the  greatest  opening  of  the, 
Compasses  ? 


GEOMETRY    FOR    BEGINNERS. 


[§87- 


E 

Fig.  82. 


87.  From  §  84  it  appears  that  every  point  in  the  perpendicu- 
lar erected  at  the  middle  of  a  line  is  equally 
distant  from  the  ends  of  the  line.  Moreover, 
every  point  not  in  this  perpendicular  is  un- 
equally distant  from  the  ends  of  the  line. 

For  let  D  (Fig.  82)  be  such  a  point,  C  E 
being  the  perpendicular.    Since  A  D  —  A  C 
+  CD  =  B  C+  C  D,  and  since  also  B  C  + 
C  D>  BD  (why?),  therefore  AD>  BD. 
Hence,  — 

Theorem.  —  Every  point  equidistant  from  the  ends  of  a  straight 
line  is  in  the  perpendicular  which  bisects  the  line. 

§  88.    Problem.  —  To  erect  a  perpendicular  at  a  given  point 
C  (Fig.  83)  of  a  given  line  A  B. 

Analysis.  —  If  we  find  in  A  B  two  points,  M  and  N,  equally  dis- 
tant from  C,  and  then  find  any  other  point  D  equidistant  from  M 

and  N,  it  is  evident  from  §  87  that 
the  line  passing  through  C  and  D 
will  be  the  perpendicular  required. 
Construction.  —  With  the  centre 
C  describe  an  arc  cutting  A  Bin  M 
and  N.  With  M  and  N  as  centres 

M  u  ~~N         and  equal  radii,  describe  arcs  inter- 

secting at  a  point  D.    Join  CD. 


/ 


83. 


Note.  —  In  practice  this  construction  is  usually  effected  by  means  of  the  ruler 
and  the  square  (see  §  49). 

Exercises.  —  1.  In  the  above  construction  what  is  the  least  value  which 
the  equal  lines  M  D  and  A7 D  can  have? 

2.    Erect  a  perpendicular  at  the  end  of  a  line  without  prolonging  the  line. 

Construction.  —  Let  A  C  (Fig.  84}  be  the  line.  Construct  on  A  C  as  base 
an  equilateral  triangle  ABC  (how  is  this  done?),  and  prolong  A  B  to  D, 
making  B  D  =  A  B.  CD  is  the  perpendicular  required. 

proof.  —  Since  A  ABC  is  equilateral,  ABC=  60°;  therefore,  in  A 
BDC=fo°  (§66).  Now  A  B  CD  is  isosceles  (why?); 


§  89.]  CHAPTER   IV. TRIANGLES.  99 

hence  B  CD  =  B  D  C-  30°.     Therefore  A  CD  ~  A  C  B  +  B  CD  =  60° 
+  30°  =  90°. 

3.    Construct   an  equilateral  triangle  having  the  f 

altitude  4Ocm.  / 

Hints.  —  Draw  a  line  40°™  long,  erect  at  one  end  / 

a  perpendicular,  at  the  other  end  make  two  angles  of  30°  / 

each. 

4-.    Construct  an  isosceles  triangle  having  given  X 

(<7)  the  base  and  the  altitude  ;    (b)  a  side  and  the 
altitude. 

5.  Bisect  the  three  sides  of  a  triangle,  and  then 
erect  perpendiculars  at  the  points  of  bisection.  In 
how  many  points  do  they  intersect?  A  Fig.  84. 


§  89.    Problem.  —  To  let  fall  a  perpendicular  from  a 
point  C  (Fig.  85}  to  a  given  line  A  B. 

The  analysis  and  the  construction 
are  the  same  as  in  the  last  problem, 
the  only  difference  being  that  in 
this  case  the  point  C  is  not  on  the 
line  A  B.  Give  the  analysis  and  the 
construction  in  full. 


Note.  —  This  construction  is  also  usually      — ****.. 

effected   in  practice   by  means  of  the  ruler     •« 

and  the  square  (§  49).  Fig.  85. 

Exercises.  —  1.    Choose  a  point  on  each  side  of  a  line,  and  let  fall  per- 
pendiculars from  each  point  to  the  line. 

2.  Draw  a  triangle,  and  from  the  vertex  of  each  angle  let  fall  a  perpen- 
dicular to  the  opposite  side.     In  how  many  points  do    the  perpendiculars 
intersect? 

3.  Through  a  given  point  draw  a  parallel  to  a  given  line  by  constructing 
two  perpendiculars. 


§  90.    Theorem.  —  Two  parallel  lines  are  everywhere  equally 
distant  from  each  other. 

Proof.  —  From  any  two  points  A  and  B  (Fig.  86)  of  one  of  the 


100  GEOMETRY    FOR    BEGINNERS.  [§  91. 

lines  let  fall  perpendiculars  meeting  the  other  line  in  C  and  D  ;  then 
A  _  B  A  C  II  B  D  (§  57,  Corollary)  .   Join 

/  B  C.      The    triangles  ABC  and 

£  CD  are  equal  (  §  56  and  I.  Law  of 
Equality  )  .  Therefore  A  C  =  B  D  ; 
that  is,  any  two  points  of  one  line 


—c 

Fig.  86.  are  equally  distant  from  the  other 

line  :   hence  the   lines   are   every- 
where equally  distant  from  each  other. 

§  91.  Let  the  problem  be  proposed  :  to  find  a  point  at  a  given 
distance  from  a  given  point. 

It  is  evident  that  any  point  in  the  circumference  of  a  circle,  de- 
scribed with  the  given  point  as  centre  and  the  given  distance  as 
radius,  will  satisfy  the  required  condition,  and  will,  therefore,  be  a 
solution  of  the  problem.  It  appears,  then,  that  there  are  an  inde- 
finite number  of  solutions  to  the  problem;  hence  the  problem  is 
termed  indeterminate. 

The  required  point  is  limited  in  position  to  a  certain  line  ;  this 
line  is  called  the  locus  of  the  point. 

Definition.  —  The  line  (or  lines)  in  which  a  point  must  be,  in 
order  to  satisfy  a  given  condition,  is  called  the  Locus  *  of  the  point. 

Again,  let  it  be  proposed  :  to  find  a  point  equidistant  from  two 
given  points. 

It  follows  from  §  87  that  the  required  point  may  be  anywhere 
in  the  perpendicular  which  bisects  the  straight  line  joining  the 
given  points.  The  problem,  therefore,  is  indeterminate,  and  the 
locus  of  the  point  required  is  the  above-mentioned  perpendicular. 

Generally  speaking,  when  a  point  has  to  satisfy  only  one  geomet- 
rical condition,  the  problem  is  indeterminate,  and  the  solution  con- 
sists in  finding  the  locus  of  the  required  point. 

1  Plural  loci. 


§  92.]  CHAPTER    IV. TRIANGLES.  10 1 

Exercises.  —  1,  In  the  cases  above  considered,  the  points  were  supposed 
to  be  confined  to  one  plane  :  what  would  the  loci  be  if  this  restriction  were 
removed  ? 

Ans.  In  the  first  case,  the  surface  of  a  sphere  with  the  given  point  as  centre  and 
the  given  distance  as  radius ;  in  the  second  case,  the  plane  perpendicular  to  the 
line  which  joins  the  given  points,  and  bisecting  it. 

2.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given  straight  line. 

3.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given  circumference. 

4.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given  plane  surface. 

5.  Find  the  locus  of  a  point  equidistant  from  two  given  parallel  lines. 

6.  Find  the  locus  of  a  point  equidistant  from  two  given  intersecting  lines. 
Hints.  —  The  locus  consists  of  the  bisectors  of  the  angles  made  by  the  lines. 

To  prove  this,  let  Pbe  any  point  on  either  of  the  bisectors,  P M  and  PN  perpen- 
diculars from  P  to  the  two  lines,  O  the  point  where  the  lines  meet;  then  show 
that  &POM^/\PON,  and  therefore  that  PM—PN.  Why  is  it  that,  although 
there  are  four  angles  to  be  bisected,  the  locus  will  consist  of  only  two  straight  lines  ? 

§  92.  When  a  point  has  to  be  found  which  satisfies  two  condi- 
tions, the  problem  (if  possible  at  all)  generally  has  one  or  two 
solutions ;  in  other  words,  the  problem  is  determinate. 

Such  problems  are  solved  by  constructing  the  loci  of  points 
which  satisfy  each  condition  separately  ;  then  the  point,  or  points, 
where  the  loci  intersect,  being  common  to  both  loci,  will  satisfy 
both  conditions,  and  will  be  the  solution  of  the  problem. 

For  example  :  find  a  point  which  shall  be  at  given  distances 
from  two  given  points. 

The  required  point  must  be  in  a  circumference  having  the  first 
given  point  for  centre  and  the  first  given  distance  for  radius  ;  and 
it  must  also  be  in  a  circumference  having  the  second  given  point 
for  centre  and  the  second  given  distance  for  radius ;  hence  it  must 
be  at  the  intersection  of  these  two  circumferences.1 

Since,  in  general,  these  circumferences  intersect  in  two  points, 
there  will  be  in  general  two  solutions  of  the  problem.  When  will 
there  be  only  one  solution  ?  When  will  there  be  no  solution  at  all  ? 

In  the  following  exercises,  after  having  given  the  solution,  state 
under  what  conditions  (if  any)  the  problem  is  impossible. 

1  This  reasoning  has  been  already  employed  in  \  75. 


102  GEOMETRY    FOR    BEGINNERS.  [§  93. 

Exercises.  —  1.  Find  a  point  in  a  given  straight  line  at  a  given  distance 
from  a  given  straight  line. 

2.  Find  a  point  in  a  given  straight  line  at  equal  distances  from  two  other 
straight  lines. 

3*  Construct  an  isosceles  triangle  having  a  given  base,  and  each  of  its 
sides  equal  to  three  times  the  base. 

4.  Find  a  point  at  a  given  distance  from  a  given  point,  and  at  the  same 
distance  from  a  given  straight  line. 

5.  Construct  a  triangle  having  given  the  base,  the  sum  of  the  other  sides, 
and  one  of  the  angles  at  the  base. 

6.  Construct  a  triangle  having  given  the  base,  the  difference  of  the  other 
sides,  and  one  of  the  angles  at  the  base. 

7.  Find  a  point  which  shall  be  equidistant  from  three  given  points. 

8.  Find  a  point  which  shall  be  equidistant  -from  three  given  intersecting 
straight  lines  (see  §  91,  Exercise  6).     Show  that  there  are  four  solutions, 
and  construct  the  four  points.     How  is  the  solution  modified  if  two  of  the 
given  lines  are  parallel? 


§  93.  The  complete  solution  of  a  problem  consists  of  three 
parts  :  — 

(*.)  The  ANALYSIS,  in  which  we  explain  how  a  correct  construc- 
tion can  be  based  upon  known  geometrical  truths. 

(it.)  The  CONSTRUCTION,  in  which,  guided  by  the  analysis,  we 
make  the  required  lines,  angles,  etc.,  with  the  help  of  the  ruler 
and  the  compasses. 

(in.)  The  DISCUSSION,  in  which  we  consider  (a)  whether  there 
may  be  more  than  one  solution ;  (b)  whether  the  data  may  have 
such  values  that  no  solution  is  possible ;  (c)  whether  the  solution 
has  any  other  peculiarities  for  particular  values  of  the  data. 

The  analysis  of  a  problem  that  can  be  solved  by  the  method  of 
loci  has  already  been  given  (§  92)  ;  in  all  other  cases  a  theorem 
(or  theorems)  must  first  be  found  on  which  a  correct  construction 
can  be  based.  In  this  search  nothing  but  experience  and  ingenu- 
ity will  ensure  success  ;  but  there  is  one  general  rule  which  will  be 
found  very  useful ;  namely  :  — 


§   93-]  CHAPTER   IV.  —  TRIANGLES.  103 

Rule.  —  Suppose  the  solution  effected,  and  draw  a  suitable  figure  ; 
then  trace  the  relations  among  the  parts  of  the  figure  until  some  re- 
lation is  discovered  which  will  give  a  clue  to  the  right  construction. 

We  will  give  an  example  to  illustrate  the  application  of  this  rule. 

Problem.  —  Through  a  given  point  to  draw  a  line  which  shall 
cut  off  equal  lengths  from  the  sides  of  a  given  angle. 

Analysis.  —  Let  P  (Fig.  87)  be  the  given  point,  B  A  C  the 
given  angle.  Suppose  that  O  P  Q  were  the  line  required ;  then  we 
know  that  A  O  Q  must  be  an  isos- 
celes triangle  (why  ?) ,  and  hence,  that 
if  we  bisected  the  angle  A,  O  P  Q 
would  be  perpendicular  to  the  bi- 
sector (§80,  Exercise  1).  This 
suggests  to  us  the  correct  construc- 
tion. 

Construction.  —  Bisect  the  angle  Fig. 

A,  let  fall  from  P  a  perpendicular  to 

the  bisector,  and  produce  it  until  it  meets  the  sides  of  the  given 
angle  in  O  and  Q  :   O  P  Q  is  the  line  required. 

Discussion.  —  Construct  the  figure  for  the  case  where  P  is  not 
between  the  sides  of  the  given  angle.  For  what  position  of  P  is 
the  solution  impossible? 

Exercises.  —  1.  Draw  a  line  which  shall  pass  through  a  given  point  and 
make  equal  angles  with  two  given  intersecting  lines. 

2.  In  a  given  straight  line  find  a  point  which  shall  be  equidistant  from 
two  given  points. 

3.  In  one  side  of  a  triangle  find  a  point  which  shall  be  equidistant  from 
the  other  two  sides  (see  §  91,  Exercise  6). 

4.  In  one  side  of  an  angle  a  point  is  given  ;   find  in  the  same  side  another 
point  which  shall  be  equidistant  from  the  first  point,  and  from  the  other  side 
of  the  angle. 

5.  From  a  given  point  without  a  given  straight  line  draw  a  line  making  a 
given  angle  with  the  given  line. 

6.  Trisect  a  right  angle  (that  is,  divide  it  into  three  equal  parts). 


104  GEOMETRY    FOR    BEGINNERS.  [§   94, 


VI.  —  Applications. 

§  94.  If  I  wish. to  find  the  distance  from  A  to  B  (Fig.  88)  it 
is  plain  that  the  intervening  pond  will  prevent  me  from  using  a 
chain  or  other  direct  means  of  measurement. 

In  numerous  instances  direct  measurement  would  be  very 
troublesome  or  quite  impossible.  On  the  ground  obstacles,  such 
as  houses,  water,  swamps,  are  often  met  with ;  if  we  want  to  meas- 
ure the  distance  from  the  earth  to  the  moon,  only  one  end  of  the 
line  to  be  measured  is  accessible  ;  while  in  the  case  of  the  distance 
between  the  sun  and  a  planet,  the  line  is  wholly  inaccessible. 

Here  Geometry  comes  to  our  aid  by  giving  us  methods  of  indi- 
rect measurement ;  in  which,  for  example,  by  measuring  one  line 
we  are  able  to  learn  the  length  of  another. 

How  this  can  be  done  by  the  help  of  the  laws  of  equal  triangles, 
we  are  now  prepared  to  understand. 

All  cases  of  the  indirect  measurement  of  a  line  may  be  reduced 
to  four,  — 

(/.)    Both  ends  of  the  line  are  accessible ; 

(«'.)   Only  one  end  is  accessible  ; 

(tit.)  Both  ends  are  inaccessible ; 

(tv.)  The  line  is  wholly  inaccessible. 

§  95.  Problem.  —  To  measure  a  line  the  ends  of  which  only 
are  accessible. 

Method  I.  —  Let  A  B  (Fig.  88)  be  the  line  to  be  measured. 
Choose  a  point  C  from  which  A  and  B  are  both  visible.  Measure 
A  C,  B  C,  and  the  angle  A  C  B. 

Prolong  A  C  to  D,  making  C  D  =  C  A,  and  prolong  B  C  to  E, 
making  C  E  =  C  B.  &DEC^&ABC  (why?).  There- 
fore DE  =  A  B,  and  its  length  (which  can  be  measured  directly) 
is  the  distance  required. 


§96-] 


CHAPTER   IV. TRIANGLES. 


105 


Method  II.  —  Measure  A  C,  B  C,  and  A  C  B,  as  before.  Then 
construct  on  paper  an  angle 
acb  —  ACB;  choose  a  suita- 
ble scale  of  reduction,  and  lay 
off  on  the  paper  the  reduced 
lengths  a  c  and  b  c  of  the  lines 
A  C  and  B  C.  Join  a  b  :  its 
length  gives,  to  the  reduced  scale, 
the  distance  A  B  required. 

For  example  :  if  A  C  =  8oom, 


Fig.  88. 


D 


B  C  —  iooom,  and  the  scale  of 

reduction  is  i  :  5000,  then  a  c  =  i6cm,  and  b  c  =  2Ocm.     If  now  we 

find  that  a  b  =  24cm,  then  A  B  =  24cm  x  5000  =  1200  metres. 

Remark.  —  The  first  method  is  preferable,  because  any  error 
made  in  measuring  a  b  is  multiplied  5000  times  in  the  result. 

How  can  the  length  of  the  pond  (Fig.  88)  be  found  ? 

§  96.    Problem.  —  To  measure  a  line  one  end  only  of  which  is 
accessible. 

Method  I.  —  Let  A  B  (Fig.  89)  be  the  line  to  be  measured,  B 
the  end  which  is  accessible. 
At  a  point  C,  in  A  B  pro- 
longed, place  a  signal  (a 
pole  or  flag).  Then  place 
a  signal  at  a  convenient 
point  D,  and  measure  the 
distances  BD  and  CD. 
Prolong  CD  to  E,  making 
DE  =  DC,  and  prolong 
B  D  to  F,  making  DF  = 
DB.  Place  signals  at  E 
and  F.  Then  proceed  in 
the  direction  EF  until  a 


106 


GEOMETRY  FOR  BEGINNERS. 


[§97- 


point  G  is  reached  which  falls  in  the  line  A  D.  Measure  F  G  : 
its  length  is  equal  to  the  distance  A  B  required. 

Proof:  A  D  E  F^D  B  C  (II.  Law  of  Equality).  Therefore 
angle  DEF  =  D  CB  ;  hence  E  G  II  A  C  (§  57,  Remark).  In 
the  triangles  DAB  and  D  F  G,  D  B  =  D  F,  ADB  =  FD  G 
(why?),  and  AB D  =  DFG  (why?)  ;  therefore  A  DAB^ 
A  DF  G  (I.  Law  of  Equality),  and  F  G  =  A  B. 

Method  II. —  (By  an  isosceles  triangle.)  Choose  a  convenient 
direction  B  H  for  running  a  straight  line  from  B,  measure  the 
angle  A  B  H,  and  find  in  B  H  a  point  H  at  which  the  angle  A  HB 
shall  be  equal  to  £  (180  —  ASH).  Then  B  H  =  AB  (why?). 
What  will  be  the  value  of  A  HB  ifABH=9o°? 

Remark.  —  In  both  methods  three  things  have  to  be  measured 
directly  :  in  Method  I.,  three  lines  ;  in  Method  II.,  one  line  and  two 
angles.  But,  if  the  means  of  measuring  an  angle  are  at  hand, 
Method  II.  is  preferable,  because  it  is  easier  to  measure  an  angle 
than  a  line. 


§  97.   Problem.  —  To  measure  a  line  when  both  ends  of  it  are 
inaccessible. 

Let  A  B  (Fig.  go)  be  the  line,  and  let  us  suppose  that  a  part 

of  the  line  between  A  and  B  is  accessible,  as  shown  in  the  Figure. 

Method  I.  —  At  a  point  C  in  A  B  erect  a  perpendicular  C  D 

(§  88),  and  take  DE  =  CD. 
At  E  also  erect  a  perpendic- 
ular FG\  then  F  G  will  be 
parallel  to  A  B  (why  ?) 

Find  in  FG  the  point  F 
which  falls  in  the  line  B  D, 
and  the  point  G  which  falls  in 
the  line  A  D.  Now  &ACD 
g*  A  ED  G  (why?),  and  A 
Fig.  90.  B  CD  ^  A  EDF  (why?). 


CHAPTER   IV. TRIANGLES. 


107 


Therefore  FE  =  B  C,  and  E  G  =  A  C ;  and  hence  FE  +  E  G 
=  BC  +  AC  =  AB,m  the  line  to  be  measured. 

Method  II. —  (By  isosceles  triangles.)  In  C  E  find  the  points 
H  and  K  from  which  the  directions  of  A  and  B  respectively  make 
with  C  E  the  angle  45°.  Then  C H -f  C K  =  AS.  Explain 
fully  why. 

§  98.  Problem.  —  To  measure  a  line  which  is  wholly  inacces- 
sible. 

Method  I.  —  Choose  a  convenient  point  C  from  which  A  and  B 
are  both  visible  ;  measure  A  C  and  B  C  as  in  §  96 ;  then  measure 
A'B'as  in  §  95. 

Method  n.  —  Choose  C  as  before ;  also  a  point  F  from  which 
A  is  visible,  and  a  point 
G  from  which  B  is  visible. 
Find  E,  the  intersection 
of  A  F  and  B  C,  and  D, 
the  intersection  of  B  G 
and  A  C.  Measure  all 
the  angles  at  C,  and  also 
the  lines  C  F,  C  E,  C  G, 
CD.  Then  represent  on 
paper  the  measured  angles  ^ 

and  lines  (the  latter  to  a 

reduced  scale) .  By  prolonging  E  F  and  D  C  till  they  meet,  we 
find  the  point  on  the  paper  which  corresponds  to  A  on  the  ground  ; 
and  by  prolonging  in  like  manner  D  G  and  E  C,  we  find  the  point 
which  corresponds  to  B.  The  length  of  A  B  on  the  paper  gives 
to  the  scale  employed  the  required  distance. 


108  GEOMETRY  FOR  BEGINNER^, 

REVIEW   OF   CHAPTER   IV. 
QUESTIONS. 

1,  Define  a  plane  figure;  a  polygon;  its  perimeter. 

2.  Define  a  triangle,  and  explain  how  it  is  named. 
8.    What  six  parts  has  every  triangle? 

4:.    Distinguish  between  exterior  and  interior  angles'* 

5.  The  sum  of  any  two  sides  of  a  triangle  is  greater  than  the  third  side. 

6.  Define  equilateral,  isosceles,  and  scalene  triangles. 

7.  Define  the  terms  base,  vertex,  and  altitude.     When  will  the  altitude  lie 
outside  the  triangle? 

8.  The  sum  of  the  angles  of  a  triangle  is  equal  to  180°.     How  can  this 
truth  be  illustrated? 

9.  Why  can  a  triangle  have  only  one  right  or  one  obtuse  angle  ? 

10.  Define  acute,  right,  and  obtuse  triangles. 

11.  If  two  angles  of  a  triangle  are  known  how  can  the  third  be  found? 

12.  In  a  right  triangle  what  is  the  sum  of  the  acute  angles? 

13.  What  is  the  complement  of  an  angle? 

14r.    An  exterior  angle  of  a  triangle  =  the  sum  of  the  opposite  interior  angles. 

15.  Define  similar,  equivalent,  and  equal  magnitudes,  and  give  examples  of 
each  kind. 

16.  In  two  triangles  what  are  corresponding  side%  (or  angles)  ? 

17.  What  is  the  least  number  of  parts  which  determine  the  size  and  shape 
of  a  triangle?     Of  these  how  many  at  least  must  be  sides? 

18.  In  what  cases  do  three  parts  fail  to  determine  the  triangle? 

19.  When  are  two  triangles  mutually  equiangular? 

20.  Construct  a  triangle,  having  given, — 

(*'.)     a  side  and  two  angles  ; 

(».)  two  sides  and  the  included  angle  ; 

(m.)  two  sides  and  the  angle  opposite  to  the  greater  side ; 

(zz/.)  the  three  sides. 

21.  State  the  four  laws  of  equal  triangles. 

22.  Make  a  triangle  equal  to  a  given  triangle. 

23.  Make  an  angle  equal  to  a  given  angle. 

24.  Draw  a  line  parallel  to  a  given  line. 

25.  What  is  a  problem  ?     What  are  auxiliary  lines,  and  how  distinguished 
from  others  in  the  construction? 


CHAPTER   IV. REVIEW,  109 

26.  In  an  isosceles  triangle  the  angles  opposite  to  the  equal  sides  are  equal. 
Corollary. 

27.  If  two  angles  of  a  triangle  are  equal  the  triangle  is  isosceles.    Corollary. 

28.  What  is  the  converse  of  a  theorem?     Give  an  example. 

29.  In  a  triangle   the  greater  of  two  angles  is  opposite  to  the  greater  side, 
and  conversely.     Corollary. 

30.  The  perpendicular  is  the  shortest  line  from  a  point  to  a  straight  line. 
Corollary. 

81.    Oblique  lines  equally  distant  from  the   foot  of  the  perpendicular  are 
equal. 

32.  Of  two  oblique  lines  unequally  distant  from  the  foot  of  the  perpendicu- 
lar, the  more  remote  is  the  greater. 

33.  If  two  isosceles  triangles  have  a  common  base  the  line  joining  their 
vertices  (z.)  bisects  the  angle  at  the  vertices,  (zY.)  bisects  the  base,  (in.) 
is  perpendicular  to  the  base. 

34:.  Bisect  a  given  angle. 

35.  Bisect  a  given  straight  line. 

36.  Every  point  equidistant  from  the  ends  of  a  straight  line  is  in  the  perpen- 
dicular bisecting  the  line. 

37.  Erect  a  perpendicular  at  a  given  point  of  a  given  line. 

38.  Let  fall  a  perpendicular  from  a  given  point  to  a  given  line. 

39.  Two  parallel  lines  are  everywhere  equally  distant  from  each  other. 

40.  Give  an  example  of  an  indeterminate  problem. 

41.  Define  the  locus  of  a  point,  and  illustrate  the  definition. 

42.  Give  an  example  of  a  determinate  problem. 

43.  How  are  problems  solved  by  the  method  of  loci? 

44.  What  are  the  three  parts  of  the  complete  solution  of  a  problem? 

45.  What  general  rule  is  often  useful  in  solving  problems? 

46.  What  four  cases  may  occur  in  the  indirect  measurement  of  a  line? 

47.  Measure  a  line  the  ends  only  of  which  are  accessible. 

48.  Measure  a  line  one  end  only  of  which  is  accessible. 

49.  Measui'e  a  line  neither  end  of  which  is  accessible. 

50.  Measure  a  line  which  is  wholly  inaccessible. 

EXERCISES. 

1.  In  an  isosceles  triangle  the  exterior  angles  at  the  base  are  equal. 

2.  In  an  isosceles  triangle  the  bisector  of  an  exterior  angle  at  the  vertex  is 
parallel  to  the  base. 


110  GEOMETRY   FOR   BEGINNERS. 

3.  In  an  isosceles  triangle  the  bisectors  of  the  base  angles  (prolonged  until 
they  meet  the  sides)  are  equal. 

4.  In  what  cases  are  we  able,  knowing  one  angle  of  a  triangle,  to  find  the 
other  two  angles? 

5.  If  in  a  right  triangle  the  acute  angles  are  30°  and  60°,  the  hypotenuse  is 
equal  to  twice  the  smaller  leg. 

6.  State  and  prove  the  converse  of  the  preceding  theorem. 

7.  Divide  an  isosceles  triangle  into  two  equal  right  triangles. 

8.  Make  an  angle  equal  to  three  times  a  given  angle. 

9.  Make  an  angle  of  15°;   also  an  angle  of  150°. 

10.  In  what  two  ways  can  an  angle  of  22^°  be  constructed? 

11.  Name  various  angles  between  o°  and  180°  which  can  be  constructed  by 
means  of  the  theorems  and  problems  already  given. 

12.  A  straight  railway  passes  within  two  miles  of  a  town.     A  place  is  de- 
scribed as  four  miles  from  the  town  and  one  mile  from  the  railway.   How 
many  places  satisfy  the  conditions? 

13.  Place  a  line  of  given  length  between  the  sides  of  an  angle  so  as  to  be 
parallel  to  a  given  line. 

14.  Through  a  given  point  between  two  lines  not  parallel  draw  a  line  which 
shall  be  bisected  at  that  point. 

15.  Three  lines  meet  in  a  point ;   draw  a  line  such  that  the  parts  of  it  inter- 
cepted by  the  three  lines  shall  be  equal. 

16.  What  axiom  is  implied  in  the  last  line  of  the  proof  of  the  theorem  in  §  87? 

17.  The  perpendiculars  erected  at  the  middle  points  of  the  sides  of  a  triangle 
meet  in  one  point.  , 

Hints,  —  Erect  two  of  the  perpendiculars ;  show  (by  II.  Law  of  Equality)  that 
their  intersection  is  equidistant  from  the  three  corners  of  the  triangle,  then 
make  use  of  §  80,  Exercise  2. 

18.  The  bisectors  of  the  three  angles  of  a  triangle  meet  in  one  point. 
Hints.  —  Draw  two  bisectors;  show  by  $  91,  Exercise  6,  that  their  intersection 

is  equidistant  from  the  sides  of  the  triangle,  and  (by  III.  Law  of  Equality), 
that  the  line  joining  the  intersection  with  the  third  corner  bisects  the  angle 
at  that  corner. 

19.  Trisect  a  given  straight  line. 

20.  Illustrate  by  an  example  the  application  of  the  rule  in  §  93. 

21.  Choose  two  stations  on  the  ground,  and  find  their  distance  by  means  of 
(i.),  §  95;  (».),  §  96;  (iii^  §  97;  <>.),  §  98. 

22.  Can  you  find  the  height  of  an  object  (a  tree,  a  church-spire,  etc.)  by 
means  of  truths  established  in  this  chapter?     How? 

230    How  many  and  what  parts  are  sufficient  to  determine  a  right  triangle?  an 
isosceles  triangle  ?  an  equilateral  triangle  ?  an  isosceles  right  triangle  ? 


99-]  CHAPTER   V.  —  QUADRILATERALS.  Ill 


CHAPTER  V. 
QUADRILATERALS. 

CONTENTS.  — I.  Sides  and  Angles  of  a  Quadrilateral  (§§  99,  100).  II.  Different 
Kinds  of  Quadrilaterals  (§§  101-105).  HI-  Construction  of  Quadrilaterals 
(§§  106-111).  IV.  Subdivision  of  a  Lane  (§§  112-114). 

J.  —  Sides  and  Angles  of  a  Quadrilateral. 

§  99.  A  plane  figure  bounded  by  four  straight  lines  is  called  a 
QUADRILATERAL  (A  B  CD,  Fig.  92). 

A  straight  line  A  C,  which  joins  two 
opposite  corners  of  a  quadrilateral,  is 
called  a  DIAGONAL. 

Exercises.  —  1.  How  many  sides  and  how 
many  angles  has  a  quadrilateral? 

2.  How  is  a  quadrilateral  named  or  de- 
noted? 

3.  What  is  the  perimeter  of  a  quadrilateral  (§  59)  ? 

4.  Into  how  many  triangles  is  a  quadrilateral  divided  by  a  diagonal? 

5.  How  many  diagonals  can  be  drawn  in  a  quadrilateral? 

§  100.  If  we  divide  a  quadrilateral  AB  CD  (Fig.  92)  into 
two  triangles  by  drawing  a  diagonal  AC,  it  is  evident  that  the  sum 
of  the  four  angles  of  the  quadrilateral  is  the  same  as  the  sum  of 
the  six  angles  of  the  two  triangles.  Now  the  sum  of  the  angles  of 
the  two  triangles  is  360°  (why?)  ;  hence,  — 

Theorem.  —  The  sum  of  the  angles  of  a  quadrilateral  is  equal 
to  four  right  angles,  or  360°. 

Exercises.  —  1.  If  the  angles  of  a  quadrilateral  are  all  equal,  what  is  the 
value  of  each? 

2.   Three  angles  of  a  quadrilateral  are  70°,  40°,  and  120°;  find  the  fourth. 


112  GEOMETRY    FOR    BEGINNERS.  [§    IOI. 

II.  — Different  Kinds  of  Quadrilaterals. 

§  101.   With  respect  to  the  position  of  the  sides  there  are  three 
kinds  of  quadrilaterals. 


A  quadrilateral  which  has  no  sides  parallel  is  called  a  TRAPEZIUM 
(Fig.  93,1.). 

A  quadrilateral  which  has  only  two  sides  parallel  is  called  a 
TRAPEZOID  (Fig  93,  II.). 

A  quadrilateral  which  has  its  opposite  sides  parallel  is  called  a 
PARALLELOGRAM  (Fig.  93,  III.). 

Exercises.  —  1.  Draw  (free-hand)  a  trapezium,  a  trapezoid,  and  a  paral- 
lelogram. 

2.    Draw  a  trapezoid  having  two  right  angles. 

§  102.  Let  AB  CD  (Fig.  94)  be  a  parallelogram.  Draw  the 
diagonal  BD.  The  alternate  angles  m 
and  n  are  equal,  and  likewise  the  alter- 
nate angles  /  and  q  (§  56).  Therefore 
A  ABD  ^  A  CBD  (I.  Law  of  Equal- 

D  c          ity),  and  AB  =  CD,  and  AD  =  BC. 

Also  the  angle  A  equals  the  angle  C ;  and, 

since  m  —  n  and  p  —  q,  therefore  m  -\-  n  =  p  -\-  q;   that  is,  the 
angles  at  B  and  D  are  equal.     Hence.  — 

Theorem  I. — Every  parallelogram  is  divided  by  a  diagonal 
into  two  equal  triangles. 


§  I03-] 


CHAPTER   V. QUADRILATERALS. 


113 


Theorem  II.  —  The  opposite  sides  of  .a  parallelogram  are  equal. 
Theorem  III.  —  The   opposite  angles  of  a  parallelogram  are 
€qual. 

Corollaries.  —  From  Theorem  II.  it  follows  that, — 

1.  Parallels  between  parallels  are  equal. 

2.  Perpendiculars  between  parallels  are  equal  (§  57,  Corol- 
lary). 

3.  Parallels  are  everywhere  equally  distant  (see  §  90) . 

Exercises.  —  1.    If  two  adjacent  sides  of  a  parallelogram  are  equal,  what 
follows  as  to  the  other  sides? 

2.  What  relation  exists  between  two  adjacent  angles  of  a  parallelogram? 
Why? 

3.  If  one  angle  of  a  parallelogram  is  a  right  angle,  what  must  the  others 
be?     Illustrate  with  a  figure. 

4»    If  one  angle  of  a  parallelogram  is  acute,  what  can  be  inferred  of  the 
other  angles?     Illustrate  with  a  figure. 

5.  One  angle  of  a  parallelogram  is  (*'.)  45°,  («'.)  54°  1 8',  (m.)  1 21°  1 6' 
44" ;   find  the  other  angles. 

6.  Prove  the  converse  of  Theorem  II. 


§  103.  The  sides  of  a  parallelogram  may  be  either  equal  or 
-unequal,  and  the  angles  may  be  either  right  or  oblique ;  so  there 
are  in  all  four  kinds  of  parallelograms :  the  oblique  unequal-sided 


IV. 


parallelogram  or  RHOMBOID  (Fig.  95,  I.),  the  oblique  equal- sided 
parallelogram  or  RHOMBUS  (Fig.  95,  II.),  the  right  unequal-sided 


114  GEOMETRY   FOR   BEGINNERS.  [§   104. 

parallelogram  or  RECTANGLE  (Fig.  95,  III.),  and  the  right  equal- 
sided  parallelogram  or  Square  (Fig.  95,  IV.). 

Exercises.  —  1.    Give  examples  of  parallelograms  (surface  of  a  table,  roof 
of  a  house,  etc.),  and  state  the  kind  in  each  case. 

2.  Draw  (free-hand)  a  parallelogram  of  each  kind. 

3.  One  angle  of  a  rhombus  is  60°;   find  the  other  angles. 

4.  One  side  of  a  rhombus  is  24™;   find  its  perimeter. 

5*   The  sides  of  a  rectangular  park  are  400'"  and  240™.     How  many  trees 
must  be  set  out  in  its  perimeter  if  they  are  to  stand  iom  apart? 

§104.    In   the   parallelogram  ABCD   (Fig.   96),    draw  the 

diagonals  AD  and  BC ;  how  many 
triangles  are  thus  formed?  A  AOB 
^  C  O  D  (why  ?)  ;  therefore  the 
sides  opposite  to  the  equal  angles 
are  equal;  that  is,  AO  =  DO  and 
BO  =  CO  :  hence,  — 

z?*          Ji  •L^ 

Theorem.  —  The  two  diagonals 
of  a  parallelogram  mutually  bisect  each  other. 

Exercises.  —  1.    Find  the  angles  which  the  diagonals  of  a  square  make 
with  the  sides  of  the  square. 

2.    In  a  square  the  diagonals  are  equal  and  perpendicular  to  each  other. 
8.    In  a  rectangle  the  diagonals  are  equal  and  inclined  to  each  other. 
Hint,  —  Draw  the  diagonals,  and  make  use  of  $  73. 

4.  In  a  rhombus  the  diagonals  are  perpendicular  to  each   other.      Are 
they  equal  or  unequal? 

5.  In    a   rhomboid  the  diagonals  are  inclined  to  each  other.     Are  they 
equal  or  unequal? 

6.  What  kind  of  parallelograms  are  divided  by  the  diagonals  into  isosce- 
les triangles?     Into  right  triangles? 


§  105.  Either  side  of  a  parallelogram  may  be  regarded  as  the 
base  ;  then  a  perpendiculai  let  fall  to  the  base  from  any  point  in  the 
opposite  side  is  the  altitude  of  the  parallelogram. 


§  io6.] 


CHAPTER   V.  —  QUADRILATERALS. 


115 


In  the  parallelogram  A  B  C  D  (Fig.  97)  A  B  is  taken  as  the 
base,  and  either  DE  or  FB  or  CG 
is  equal  to  the  altitude.     Why  are 
.DE,  FB,  and  CD  all  equal? 

In  a  rectangle  whatever  side  be 
taken  as  base  the  adjacent  side  is 
the  altitude. 

In  a  square  the  base  and  altitude  are  equal. 

In  a  trapezoid  one  of  the  parallel  sides  is  taken  as  the  base. 

In  a  trapezium  the  terms  base  and  altitude  are  not  used. 

III.  —  Construction  of  Quadrilaterals. 

§  106.   Problem.  —  To  construct  a  square,  having  given  one 
side,  a  (Fig.  98). 

Construction.  —  At  A  make  a  right  angle  ;  take  AB  =  AD  =  a't 

then  from  B  and  D  as  centres,  with  a  radius  a 

equal  to  a,  describe  arcs  intersecting  at  a  point 
C.     AB  CD  is  the  square  required. 

Proof  1  The  four  sides  are  each  equal  to  a 
by  construction.  Join  B  D,  and  make  use  of 
§  102,  Theorem  I. 

Construct  another  square  with  the  same  side 
a  •  then  prove  (by  division  into  triangles,  and 
IV.  Law  of  Equality)  that  the  two  squares  are  equal. 

Hence,  a  square  is  completely  determined  if  one  side  is  known  ; 
and,  — 

Theorem.  —  Two  squares  are  equal  if  they  have  a  side  equal. 

Exercises.  —  1.    Construct  a  square  whose  side  shall  be  idm. 

2.  Construct  a  square  whose  perimeter  shall  be  im. 

3.  Draw  a  rectangle  ;   then  construct  a  square  with  the  same  perimeter  as 
that  of  the  rectangle. 


Fig.  98. 


1  If  the  construction  of  a  problem  is  given  without  a  previous  analysis,  a  proof 
of  the  construction  (unless  it  is  quite  obvious  without  it)  ought  to  follow. 


116  GEOMETRY   FOR   BEGINNERS.  [§   107. 

4r»   Construct  a  square,  having  given  its  diagonal. 

5.  Construct  a  square,  and  upon  each  side  an  equilateral  triangle. 

6.  Construct  an  equilateral  triangle,  and  upon  each  of  its  sides  a  square. 

§  107.   Problem.  —  To  construct  a  rectangle,  having  given  two 
adjacent  sides,  a  and  b  (Fig.  99) . 

Construction  and  proof  similar  to  those  of  the  last  section. 

Construct    another   rectangle   with 

~~~b  the   same  sides,  and  prove  that  the 

two  rectangles  are  equal. 

Hence,  a  rectangle  is  determined  if 
we  know  two  adjacent  sides  ;  also,  — 
Theorem.  —  Two   rectangles   are 
equal  if  they  have  tivo  adjacent  sides 
equal  each  to  each. 

Exercises. — 1.    Construct  a  rectangle  with  the  sides  4Ocm  and  6ocm.     (If 
on  paper,  use  a  reduced  scale.) 

2.  Make  a  plan  of  a  rectangular  field  whose  sides  measure  640™  and  360™. 

3.  How  many  rectangles  can  be  constructed  upon  a  given  line  as  the 
diagonal? 

§  108.   Problem.  —  To  construct  a  parallelogram,  having  given 
two  adjacent  sides  and  the  included  angle. 

Construction.  —  Let  a  and  b  (Fig.  100)  be  the  given  sides,  70° 

_a the  included  angle.     Make  an  angle 

A  =  70°,  then  proceed  as  in  the 
last  section.  A  B  C  D  is  the  required 
parallelogram. 

Proof  similar  to  that  in  §  106. 
Construct  a  second  parallelogram 
with  the  same  given  parts,  and  prove 
that  it  is  equal  to  the  first. 

Therefore,  tivo  sides  and  the  in- 
Fig.  ioo.  eluded  angle  completely  determine  a 

parallelogram  ;  and,  further,  — 


§  lOp.]  CHAPTER  V.  —  QUADRILATERALS.  117 

Theorem.  —  Two  parallelograms  are  equal  if  two  sides  and 
the  included  angle  of  one  are  equal  respectively  to  two  sides  and 
tlce  included  angle  of  the  other. 

Exercises.  —  1.  Construct  a  parallelogram,  the  two  sides  being  44cm  and 
66cm,  and  the  included  angle  1 10°. 

2.  Construct  a  rectangle,  having  given  — 

(a)  A  side  and  a  diagonal ; 

(b}   A  side  and  the  opposite  angle  of  the  diagonals ; 

(c)   A  diagonal  and  the  angle  between  the  diagonals. 

3.  Construct  a  rhombus,  having  given  — 

(a)  A  side  and  an  angle  ; 

(b)  A  side  and  a  diagonal ; 
(r)    The  diagonals  ; 

(^/)  An  angle  and  the  diagonal  through  its  vertex. 

4.  Construct  a  rhombus  whose  angles  shall  have  the  ratio  1 : 3. 

§  109.  Problem.  —  To  construct  a  trapezoid,  having  given 
three  sides  (two  being  the  parallel  sides}  and  one  angle. 

Construction.  —  Let  a,  b,  c  {Fig.          a 

JQI)   be  the   given   sides,   68°  the 
given  angle. 

Make  an  angle  A  =  68°  ;  take  AB 
=  a,  A  D  =  b  •  through  D  draw  a 
line  parallel  to  A  B,  and  on  it  take 
D  C  =  c.  Join  B  C.  A  B  C  D  is 
the  trapezoid  required. 

Exercises.  —  1.    How  many  and  what 

parts  determine  a  trapezoid?  A  r-,.  B 

rig.  101. 
2.    Construct  a  trapezoid  with  the  sides 

6ocm,  7Ocm,  8ocm,  and  the  angle  between  the  first  two  sides  equal  to  75°. 
3*    Construct  a  trapezoid,  having  given  — 

(a)  The  two  parallel  sides  and  the  altitude  ; 
(3)    The  two  sides  not  parallel  and  the  altitude  ; 

(c)  The  two  parallel  sides  and  one  of  the  other  sides ; 

(</)  One  of  the  parallel  sides  and  the  two  sides  not  parallel. 


118  GEOMETRY   FOR   BEGINNERS.  [§   IIO. 

§  110.  Problem.  —  To  construct  a  trapezium,  having  given 
three  sides  and  the  two  angles  included  by  these  sides. 

Construction.  —  If  a,  b,  and  c  are  the  given  sides,  60°  and  75° 
the  included  angles,  draw  a  line  A  B  =  b ;  through  A  and  B  draw 
lines  making,  with  B  B,  the  angles  60°  and  75°  respectively;  lay 
off  on  these  lines  the  lengths  A  D  —  a,  and  B  C  =  c,  and  join  the 
points  thus  found. 

Draw  a  figure  to  illustrate  this  construction. 

Exercises.  —  1.  Construct  a  trapezium  with  the  sides  i2Ocm,  90°™,  70°™, 
and  the  angles  67°  and  83°. 

2.    Construct  a  trapezium,  having  given  — 

(a)  Three  angles  and  the   two   sides  between  the  first  and  third 

angles  ; 

(£)   Four  sides  and  one  angle  ; 
(<:)    Four  sides  and  a  diagonal. 

§  111.  Problem.  —  To  construct  a  quadrilateral  equal  to  a 
given  quadrilateral  A  B  CD  (Fig.  102). 

Analysis.  —  If  we  draw  the  diagonal  B  D,  we  divide  the  given 
quadrilateral  into  two  triangles.  If  now  we  construct  (§  76)  two 
new  triangles  equal  respectively  to  these,  and  similarly  placed,  it  is 

evident  that  they  will  form,  taken  to- 
gether, a  new  quadrilateral  equal  to 
the  given  quadrilateral  (Axiom  II.). 

Construction.  —  Take  E  F  =  A  B  ; 
from  E  and  F  as  centres,  with  the 
radii  AD  and  B  D  respectively,  de- 
scribe arcs  intersecting  at  a  point  H '  • 

Fig.  IO2. 

then  from  F  and  H  as  centres,  with 

B  C  and  D  C  respectively  as  radii,  describe  arcs  intersecting  at  a 
point  G.    Lastly,  join  E  H,  H  G,  and  G  F. 

Exercises.  —  1.    Draw  a  trapezium,  and  then  construct  another  equal  to  it. 
2.    When  are  two  quadrilaterals  equal  to  each  other? 


§   112.]  CHAPTER   V.  —  QUADRILATERALS.  119 


IV.  — Subdivision  of  a  Line. 

§  112.  Let  a,  b,  c,  d,  e  (Fig.  103)  be  parallel  lines  equidistant 
from  one  another,  and  let  them  be  cut  by  any  other  lines  as  in  the 
figure. 

The  perpendiculars  A  F,  B  G,  C H,  D K  are  equal  (why?). 
&ABF  ^  A  B  C  G  ^  ., 

4 M 

A  C  DH  a*  A  D  E  K 
(why  ?)  ;  therefore  A  B  = 
BC=CD  =  DE. 

In  like  manner  we  can 
prove  that  the  parts  cut 
from  any  other  line,  M ' N 
or  P  Q,  are  equal  to  each 
other.  Hence,  — 


Theorem.  —  Equidis-     I  J  \        IQ 

tant parallel  lines  cut  equal  pig.  I0j. 

parts  from  any  line  that  intersects  them. 

Exercise.  —  In  a  right  triangle  the  middle  point  of  the  hypotenuse  is  equi- 
distant from  the  three  corners  of  the  triangle. 

Hint.  —  Through  this  middle  point  draw  a  line  parallel  to  one  of  the  legs. 

§  113.   Problem.  —  To  divide  a  straight  line  A  B  (Fig.  104) 
into  any  number  (say  five}  equal  parts. 

Construction.  —  Through  A  draw  any  straight  line  A  X,  on  which 
lay  off  five  equal  parts  of  any 
length.  Join  C,  the  last  point 
thus  obtained,  to  B,  and 
through  the  other  points  of 
division  of  A  C  draw  lines 
parallel  to  B  C.  They  will 
divide  AB  into  five  equal 
parts. 


120 


GEOMETRY    FOR   BEGINNERS. 


[§ 


Proof.  —  Make  use  of  the  theorem  in  §  112. 

Exercise.  —  Divide  a  straight  line  into  3,  6,  7,  9,  10  equal  parts. 

§  114.  Problem.  —  To  divide  a  straight  line  A  B  (Fig.  705) 
into  two  parts  which  shall  have  a  given  ratio  (say  3:4). 

Construction.  —  Draw  through  A  any  line  A  C,  and  lay  off  on 

it  3  -f-  4  =  7  equal  parts.  Join  the 
last  point  C  to  B,  and  draw  through 
D,  the  third  point  of  division  from 
A,  a  line  DE  parallel  to  B  C ;  it 
intersects  AB  in  E.  Then, 


E 
Fig.  105. 


B 


Proof.  —  Lines  drawn  through  the 
other  points  of  division  of  A  C,  and  parallel  to  B  C,  would  (by 
§  112)  divide  A  E  into  three  equal  parts,  and  E  B  into  four 
equal  parts. 

Exercises.  —  1.    Divide  a  line  in  the  ratio  4 :  7. 

2.  Construct  a  rectangle,  having  given  that  the  sum  of  two  adjacent  sides 
is  i6m,  and  their  ratio  2:3. 

3.  Construct  a  rhombus  in  which  the  sum  of  the  diagonals  shall  be  I2m, 
and  one  of  them  shall  be  twice  the  other. 

4.  The  perimeter  of  a  rectangular  garden  is  8o.4m,  and  the  longer  side  is 
to  the  shorter  as  5:7.     Make  a  plan  of  the  garden,  and  find  the  lengths  of  its 
sides. 

5.  Divide  a  line  into  three  parts  which  shall  be  to  each  other  as  2 :  3 :  5. 
Hint.  —  Lay  off  on  the  auxiliary  line  2  +  3  +  5  =  10  equal  parts. 

6.  Construct  a  triangle  whose  sides  shall  be  to  each  other  as  the  numbers 
3,  4,  5.     Is  this  problem  determinate  or  not? 

7.  Construct  an  isosceles  triangle  whose  base  shall  be  to  one  of  the  equal 
sides  as  3 : 4. 

8.  Divide  a  line  into  four  parts  which  shall  be  to  each  other  as  the  num- 
bers i,  3,  4,  6. 


CHAPTER   V. REVIEW.  121 

REVIEW   OF   CHAPTER  V. 
SYNOPSIS. 

1»    A  quadrilateral  is  a  four-sided  polygon. 

2.  The  sum  of  its  angles  equals  360°. 

3.  There  are  three  classes  of  quadrilaterals :  — 

(z.)     The  trapezium  (no  parallel  sides); 
(z'z.)    The  trapezoid  (two  parallel  sides)  ; 
(z'z'z.)  The  parallelogram  (both  pairs  of  sides  parallel). 
4»    There  are  four  kinds  of  parallelograms  :  — 

(z.)     The  rhomboid  (adjacent  sides  unequal,  angles  oblique); 

(z'z.)    The  rhombus  (sides  all  equal)  ; 

(z'z'z.)  The  rectangle  (angles  all  right  angles) ; 

(zV.)  The  square  (sides  equal,  and  angles  right  angles). 

5.  A  diagonal  divides  a  parallelogram  into  two  equal  triangles.. 

6.  Hence,  the  opposite  sides  and  also  the  opposite  angles  of  a  parallelo- 
gram are  equal. 

7.  The  two  diagonals  of  a  parallelogram  mutually  bisect  each  other,  and 
are  equal  to  each  other  in  the  rectangle  and  the  square,  perpendicular 
to  each  other  in  the  rhombus  and  the  square. 

8.  Two  squares  are  equal  if  they  have  a  side  equal ;   two  rectangles,  if  their 
adjacent  sides  are  respectively  equal ;   any  two  parallelograms,  if  they 
have  two  sides  and  the  included  angle  equal  each  to  each. 

9.  Equidistant  parallel  lines  cut  equal  parts  from  any  line  that  intersects 
them. 

10.    No.  9  enables  us  to  divide  a  line  into  equal  parts,  or  into  parts  which 
shall  be  to  each  other  in  given  ratios. 

EXERCISES. 

1.  Divide  a  quadrilateral  into  a  parallelogram  and  a  triangle. 

2.  Divide  a  parallelogram  into  two  equal  triangles. 

3*    Divide  a  parallelogram  into  two  equal  parallelograms. 

4.  Divide  a  square  into  four  equal  isosceles  triangles. 

5.  Divide  a  rhombus  into  four  equal  right  triangles. 

6.  Divide  a  trapezoid  in  which  the  non-parallel  sides  are  equal  into  a  par- 
allelogram and  an  isosceles  triangk. 


122  GEOMETRY    FOR    BEGINNERS. 

7.  Divide  a  right  triangle  into  two  isosceles  triangles. 

8.  The  adjacent  angles  of  a  parallelogram  are  supplementary. 

9.  In  a  rhomboid  one  angle  is  120°;   find  the  other  angles. 

10.  If  two  opposite  sides  of  a  quadrilateral  are  equal  and  parallel,  the  figure 
is  a  parallelogram.  (Draw  a  diagonal.  Use  §§  56,  57,  73.) 

11*  The  middle  points  of  the  four  sides  of  a  parallelogram  are  the  corners 
of  a  new  parallelogram.  If  the  first  parallelogram  is  a  rectangle  the 
second  is  a  rhombus;  if  the  first  is  a  rhombus  the  second  is  a  rectangle  ; 
if  the  first  is  a  square  the  second  is  a  square. 

12.  The  bisectors  of  the  angles  of  a  parallelogram  enclose  a  rectangle.     (Ex- 
ercise 8  and  §  64.) 

13.  Prove  the  converse  of  §  102,  Theorem  III. 

14.  The  three  perpendiculars  let  fall  from  the  corners  of  a  triangle  meet  in 
one  point. 

Hints.  —  Through  the  corners  of  the  triangle  draw  lines  parallel  to  the  oppo- 
site sides,  and  prolong  them  till  they  intersect,  forming  a  new  triangle. 
Then  show  (by  means  of  \  102,  Theorem  II.,  Axiom  I.,  and  §  56,  Corol- 
lary) that  the  three  altitudes  of  the  given  triangle  are  perpendicular  to  the 
middle  points  of  the  new  triangle,  and  make  use  of  Exercise  17,  page  no. 

15.  Construct  a  rectangle,  having  given  one  side  and  the  perimeter. 

16.  Construct  a  rhombus  which  shall  have  its  obtuse  angles  twice  as  large  as 
those  which  are  acute. 

17.  Construct  a  rhomboid,  having  given  — 

(a)  Two  unequal  sides  and  the  included  angle ; 
(£)  Two  unequal  sides  and  a  diagonal ; 
(c)   One  side  and  the  two  diagonals. 

18.  Place  a  line  of  given  length  between  two  given  parallel  lines  so  that  it 
shall   pass   through  a  given  point.      (Three  different  positions  of  the 
point.) 

19.  Place  a  square  within  a  given  square  so  that  one  corner  of  it  shall  be  at 
a  given  point  in  a  side  of  the  given  square. 

20.  In  a  square  construct  an  equilateral  triangle  having  one  corner  common 
with  a  corner  of  the  square,  and  the  others  in  the  sides  of  the  square. 

21.  How  many  and  what  parts  (sides  or  angles)  are  required  to  determine  a 
square  ?    a  rectangle  ?   a  rhombus  ?   a  parallelogram  ?    a  trapezoid  ?    a 
trapezium  ? 

22.  Divide  a  line  into  3^  parts. 

23.  Construct  a  rectangle  whose  perimeter  shall  be  im,  and  the  ratio  of  the 
base  and  altitude  3 :  7.     What  are  the  lengths  of  its  sides? 


CHAPTER   VI.  —  POLYGONS. 


123 


CHAPTER  VI. 
POLYGONS. 

CONTENTS.  — I.  Sides  and  Angles  of  a  Polygon  (§§115-117).    II.  Regular  Poly- 
gons (§§  118-122).     III.  Symmetrical  Figures  (§  123). 

J.— Sides  and  Angles  of  a  Polygon. 

§  115.   A  POLYGON  is  a  plane  figure  bounded  by  straight  lines. 
What  are  its  sides  and  \\&  perimeter  ?  (§  59). 


Fig.  106. 

As  regards  the  number  of  sides,  polygons  begin  with  three 
(§  60),  and  go  on  increasing  without  limit.  As  the  number  of 
sides  increases,  the  polygon  approaches  a  circle  in  shape  (see  Fig. 
106)  •  in  other  words,  the  circle  is  the  limit  which  the  polygon 
approaches  (but  never  reaches)  when  the  number  of  its  sides  is 
increased  more  and  more. 


124  GEOMETRY    FOR    BEGINNERS.  [§   >  r6. 

The  triangle  and  the  quadrilateral  are  polygons ;  but,  as  they 
have  been  already  studied,  we  shall  now  apply  the  term  chiefly  to 
those  figures  which  have  more  than  four  sides.  A  polygon  of  five 
sides  is  called  a  PENTAGON,  of  six  sides  a  HEXAGON,  of  seven  sides 
a  HEPTAGON,  of  eight  sides  an  OCTAGON,  of  nine  sides  a  NONAGON, 
of  ten  sides  a  DECAGON,  of  twelve  sides  a  DODECAGON,  of  twenty- 
sides  an  ICOSAGON. 

Which  of  these  polygons  are  represented  in  Fig.  106  ? 

Every  polygon  has  as  many  angles  as  sides ;  each  side  has  two 
adjacent  angles ;  each  angle  is  formed  by  two  intersecting  sides. 

A  line  joining  two  corners  not  in  the  same  side  is  called  a  DIAG- 
ONAL. 

By  drawing  as  many  diagonals  as  possible  from  one  corner  of  a 
polygon  we  divide  the  polygon  into  triangles  ;  and  Exercise  i,  below, 
leads  to  the  following  general  law  :  — 

The  number  of  diagonals  which  can  be  drawn  from  one  corner 
of  a  polygon  is  THREE  less  than  the  number  of  sides  ;  and  the  num- 
ber of  triangles  into  which  the  polygon  is  divided  is  TWO  less  than 
the  number  of  sides. 

Exercises.  —  1.  Draw  polygons  of  3,  4,  5,  6,  7,  8  sides,  and  from  a  corner 
of  each  all  the  diagonals  possible  ;  then  prepare  three  vertical  columns,  in 
the  first  writing  the  number  of  sides  of  the  several  polygons,  in  the  second 
the  number  of  diagonals,  in  the  third  the  number  of  triangles.  What  general 
conclusions  follow  from  the  results? 

2.  How  many  diagonals  can  be  drawn  from  one  corner  of  a  polygon  of 
60  sides,  and  how  many  triangles  will  be  formed? 

3*    Divide  an  octagon  into  three  quadrilaterals  and  a  triangle. 

4.  If  you  cut  off  a  triangle  from  a  hexagon  with  a  diagonal,  what  kind  of 
polygon  is. left? 

5.  How  many  diagonals  in  #//can  be  drawn  in  a  quadrilateral?  a  penta- 
gon? a  hexagon?  a  decagon? 

§  116.  The  angles  of  a  polygon  may  be  acute,  right,  obtuse,  or 
even  convex.  In  the  last  case,  they  are  usually  termed  reentrant 


§   1 1 6.]  CHAPTER  VI. — POLYGONS.  125 

angles.     In  Fig.  107,  A  is  a  polygon  having  one  reentrant  angle, 
and  B  a  polygon  having  two  reentrant  angles. 


Fig.  207. 

But  whatever  be  the  values  of  the  angles,  their  sum  in  any  poly- 
gon always  bears  a  simple  relation  to  the  number  of  sides.  If  a 
polygon  could  be  divided  into  as  many  triangles  as  it  has  sides, 
then  the  sum  of  all  its  angles  would  evidently  be  equal  to  twice  as 
many  right  angles  as  there  are  sides  (§  64).  But  since  by  the 
last  section  it  appears  that  the  number  of  these  triangles  is  two  less 
than  the  number  of  sides,  it  follows  that  the  sum  of  all  the  angles 
will  be  twice  as  many  right  angles  as  there  are  sides  less  two. 
Hence,  — 

Theorem.  —  The  sum  of  the  angles  of  a  polygon  is  equal  to 
twice  as  many  right  angles  as  there  are  sides  less  two. 

Thus  the  hexagon  (Fig.  108)  consists  of  6  —  2  =  4  triangles ; 
and  the  sum  of  its  angles  =  4X2  =  8  right 
angles  =  720°. 

In  general,  if  n  be  the  number  of  sides  (n 
being  any  number),  the  following  equation 
or  FORMULA  enables  us  to  find  the  sum  of 
the  angles  :  — 

Sum  of  the  angles  (in  degrees)  =  180(n-2).  [i.]  '  p. 

Exercises.  —  1.    What  polygon  cannot  have  reentrant  angles? 
2.    Find  the  sum  of  the  angles  of  polygons  of  3,  4,  5,  6,  8,  10,  12,  20  sides. 
3*    Find  by  division  into  triangles  the  sum  of  the  angles  of  the  polygon  A 
{Fig.  707).     Also  find  the  sum  by  means  of  Formula  [i]. 


126  GEOMETRY   FOR   BEGINNERS.  [§   117. 

4.  Find  in  the  same  two  ways  the  sum  of  the  angles  of  the  polygon  B 
(Fig.  707). 

5.  Of  how  few  sides  can  you  make  a  polygon  with  one  reentrant  angle? 
with  two  reentrant  angles? 

6.  Make  a  hexagon  with  as  many  reentrant  angles  as  possible. 

7.  Make  a  hexagon  having  only  right  angles  or  their  multiples. 

8.  Draw  a  polygon  such  that  one  of  its  diagonals  shall  lie  wholly  outside 
the  polygon. 

§  117.  Problem.  —  To  make  a  polygon  equal  to  a  given  poly- 
gon ABCDEF  (Fig.  /op). 

Analysis.  —  Divide  the  given  polygon  into  triangles  by  drawing 

diagonals  from  one  corner ; 
then  construct  a  series  of 
new  triangles  equal  respec- 
tively  to  those  of  the  given 
polygon,  and  similarly 

placed.     It  is  evident  that 
Fig.  log. 

these    taken   together  will 
form  a  new  polygon  equal  to  the  given  polygon  (Axiom  II.). 
Draw  the  figures  and  give  the  construction  in  full. 

Exercises.  —  1.    Construct  a  pentagon  equal  to  a  given  pentagon. 

2.  Construct  an  octagon  equal  to  a  given  octagon. 

3.  When  are  two  polygons  equal  to  each  other? 

II.— Regular  Polygons. 

§118.  Definitions. — A  polygon  which  has  equal  sides  is 
termed  EQUILATERAL  (Fig.  no,  A)  ;  a  polygon  which  has  equal 
angles  is  termed  EQUIANGULAR  (Fig.  no,  B}  ;  and  a  polygon  which 
has  equal  sides  and  equal  angles  (Fig.  no,  C)  is  termed  REGULAR. 

Since  the  angles  of  a  regular  polygon  are  equal,  each  angle  is 
equal  to  the  sum  of  all  the  angles  divided  by  their  number ;  or,  in 
other  words,  divided  by  the  number  of  sides  of  the  polygon. 

If  we  let  n  denote  the  number  of  sides,  the  sum  of  the  angles 


§    1 1 9.]  CHAPTER    VI.  —  POLYGONS.  127 

(by  §  116)  =  1 80  (n  —  2)  =  180  w  —  360.     Dividing  this  value 
by  n  we  obtain  (Axiom  V.)  the  formula, — 

RAA 

Each  angle  of  a  regular  polygon  =  180  -— .  [2.] 


Fig.  no. 

Exercises.  —  1,  What  is  the  regular  triangle  called?  the  regular  quadri- 
lateral? 

2.    Can  a  triangle  be  equilateral  without  being  equiangular? 

3*  What  quadrilateral  is  equilateral  but  not  equiangular?  equiangular  but 
not  equilateral? 

4.  Make  three  hexagons :  the  first  equilateral  but  not  equiangular ;  the 
second  equiangular  but  not  equilatei'al ;   the  third  regular. 

5.  Find  each  angle  of  a  regular  polygon  of  3,  4,  5,  6,  8,  10,  12,  20  sides, 
and  arrange  the  results  in  a  vertical  column.    As  the  number  of  sides  increases 
how  does  the  angle  change? 

6.  Find  the  angle  of  a  regular  polygon  of  360  sides. 

7.  Express  Formula  [2]  in  words. 

8.  One  side  of  a  regular  dodecagon  is   i8m;   find  its  perimeter.     Give  a 
rule  for  finding  the  perimeter  of  a  regular  polygon  when  a  side  is  known. 

§  119.  Regular  polygons  possess  a  very  important  property 
stated  in  the  following  theorem  :  — 

Theorem.  —  The  bisectors  of  the  angles  of  a  regular  polygon 
meet  in  a  point  equally  distant  (i.)  from  all  the  corners,  and  also 
(ii.)  from  all  the  sides. 

Proof.  —  Let  the  bisectors  AO,  BO  (Fig.  in  or  112)  of  the 
angles  A,  B,  meet  in  <9,  and  join  OC,  OD,  etc.  The  triangles 
AOB,  BOC  are  isosceles  (§  81);  therefore  OA=OB.  The 


128 


GEOMETRY   FOR   BEGINNERS. 


[§  "9- 


same  two  triangles  are  also  equal  (§  73)  ;  therefore  OC=  OA. 
Also,  OCB=  OAB  —  \  one  of  the  equal  angles  of  the  polygon; 
therefore  OC  bisects  the  angle  BCD.  In  like  manner  we  can 
prove  that  O£>=  OA,  and  bisects  the  angle  CDE ;  etc. 

D 


A          G          B 
Fig.  in. 


A  G 

Fig.  JI2. 


(ii.)  The  distances  from  O  to  the  sides  of  the  polygon  are  the 
perpendiculars  let  fall  from  O  to  the  sides  (§  83,  Corollary) .  Take 
any  two  of  these  perpendiculars  OG  and  OH;  they  are  equal, 
because  A  OGB  ^  OHB  (why?).  In  like  manner  we  can 
show  that  OH  —  OJ  =  OK,  etc. ;  in  other  words,  these  perpen- 
diculars are  all  equal. 

Definitions.  —  This  remarkable  point  O  in  a  regular  polygon 
is  called  its  CENTRE  ;  the  distance  from  the  centre  to  a  corner  is 
called  the  GREATER  RADIUS  ;  the  distance  from  the  centre  to  a  side 
is  called  the  LESS  RADIUS  ;  the  angle  between  two  radii  is  called 
the  ANGLE  AT  THE  CENTRE. 

Corollaries.  —  I.  Every  regular  polygon  can  be  divided  into  as 
many  equal  isosceles  triangles  as  it  has  sides  (how  ?) . 

2.  Every  regular  polygon  can  be  divided  into  twice  as  many 
equal  right  triangles  as  it  has  sides  (how  ?) . 

3.  The  greater  radii  of  a  regular  polygon  are  all  equal,  the  less 
radii  are  equal,  and  the  angles  at  the  centre  are  equal. 

4.  The  angle  at  the  centre  is  equal  to  360  divided  by  the  num- 
ber of  sides  ;   in  other  words,  to  the  last  term  </  [2],  §  118. 


§    120.] 


CHAPTER   VI.  —  POLYGONS. 


129 


Exercises.  —  1.  Prove  the  preceding  theorem  for  the  case  of  a  triangle. 
Draw  a  figure  and  give  the  proof  in  full.  (Compare  this  exercise  with  Exer- 
cises 7  and  8,  §  92.) 

2.  Show  that  every  regular  polygon  is  composed  of  an  even  number  of 
equal  right  triangles. 

3.  Prove  that  in  a  regular  polygon  a  radius,  if  prolonged,  divides  the  poly- 
gon  into  two  equal  parts. 

4.  Prove  that  the  angle  between  two  less  radii  is  equal  to  the  angle  be- 
tween two  greater  radii.     What  value  has  the  angle  between  a  greater  radius 
and  a  less  radius? 

6.  Find  the  angle  at  the  centre  in  an  equilateral  triangle?  a  square?  a 
pentagon?  a  hexagon?  an  octagon?  a  decagon? 

6.  If  the  angle  at  the  centre  is  10°,  how  many  sides  has  the  polygon,  and 
what  is  the  value  of  each  angle? 

7.  If  an  angle  at  the  centre  is  given,  how  can  you  find  an  angle  of  the 
polygon? 

8.  If  an  angle  of  the  polygon  is  given,  how  can  you  find  an  angle  at  the 
centre? 

NOTE.  —  For  the  sake  of  reference,  the  chief  data  respecting  the  angles  of 
regular  polygons  are  collected  in  the  following  table  :  — 


NUMBER   OF   SIDES. 

SL'.M    OF   ANGLES. 

EACH   ANGLE. 

ANGLE   AT  THE 
CENTRE. 

3 

180° 

60° 

120' 

4 

360° 

90° 

90° 

5 

54°° 

108° 

72* 

6 

720° 

120° 

60° 

8 

I080° 

135° 

45° 

10 

I4400 

144° 

36° 

12 

I800° 

150° 

30° 

20 

3240° 

I620      ' 

18° 

§  120.  Problem.  —  To  construct  a  regular  polygon  having 
given  the  number  of  sides. 

Construction.  —  Construct  about  a  point  as  many  angles  as  the 
polygon  has  sides,  each  angle  equal  to  the  angle  at  the  centre  (how 
is  this  angle  found  ?)  ;  lay  off  equal  lengths  on  the  sides  of  these 
angles,  and  join  the  points  thus  determined. 


130  GEOMETRY    FOR    BEGINNERS.  [§    121. 

Exercises.  —  1.    Is  the  above  problem  determinate  or  indeterminate  ? 

2.  Construct  (z.)  an  equilateral  triangle  ;  (zY.)  a  square. 

3.  Construct  a  regular  hexagon.     Join  the  alternate  corners  by  straight 
lines.     What  figure  do  these  lines  enclose?     Why? 

4.  Prove  that  the  greater  radius  of  a  regular  hexagon  is  equal  to  a  side  of 
the  hexagon. 

5.  Construct  a  regular  octagon.     Join  the  alternate  corners   by  straight 
lines.     What  figure  do  these  lines  enclose?     Why? 

6.  Construct  a  regular  nonagon. 

7.  Can  you  devise  another  way  to  construct  a  regular  polygon? 

§  121.  Problem.  —  To  find  the  centre  of  a  given  regular  poly- 
gon. 

Analysis.  —  Two  methods  of  solving  this  problem  are  suggested 
by  §  118 ;  what  are  they  ? 

A  third  method  is  based  on  the  fact  (see  Figs,  in  and  112) 
that,  if  the  polygon  has  an  even  number  of  sides,  a  line  through  one 
corner  and  the  centre  passes  through  the  opposite  corner ;  while 
if  it  has  an  odd  number  of  sides,  this  line  passes  through  the 
middle  of  the  opposite  side.  Hence,  state  rules  for  proceeding  in 
each  case. 

Exercises.  —  1.  Find  the  centre  of  (*'.)  an  equilateral  triangle;  (M.)  a 
square  ;  (m.)  a  regular  pentagon  ;  (iv.~)  a  regular  hexagon. 

2.    Divide  an  equilateral  triangle  into  3  equal  parts. 

3»  Divide  a  sxquare  into  4  equal  parts.  Can  you  do  this  in  more  than 
one  way? 

4.  Divide  a  regular  hexagon  into  6  equal  parts  ;   also  into  12  equal  parts. 

5.  Divide  a  regular  octagon  into  4,  8,  16  equal  parts. 

NOTE.  —  The  construction,  etc.,  of  regular  polygons  will  be  considered  further 
in  the  chapter  on  the  Circle. 

§  122.  Applications. — The  stone  pavements  of  large  vesti- 
bules, galleries,  etc.,  and  the  patterns  used  in  quilting,  are  illustra- 
tions of  the  practical  employment  of  regular  polygons. 

Not  every  regular  polygon  can  be  employed  for  these  purposes ; 
only  those  can  be  used  whose  sides  touch  at  all  points ;  for,  if  this 


§  122.] 


CHAPTER  VI. — POLYGONS. 


131 


were  not  the  case,  the  surface  would  not  be  everywhere  covered. 
We  meet  most  commonly  with  groups  of  squares,  hexagons,  or 
octagons.  Let  us  examine  some  cases. 

1.  Equilateral  triangles   (Fig.  113)   can  be  used.     For,  since 
each  angle  is  equal  to  60°,  it  is  obvious  that  the  triangles  may  be 
arranged  in  groups  of  six  about  a  point,  each  group  constituting 
a  regular  hexagon. 

2.  The  arrangement  in  squares  (like  a  checker-board)  is  shown 
in  Fig.  114. 


Fig.  113. 


Fig.  if 4. 


Fig.  115. 


3.  Regular  hexagons  (Fig.  115}  may  be  arranged  in  groups  of 
three  about  a  point ;  for  the  angle  of  the  hexagon  being  120°,  three 
angles  =  3  X  120°  =  360°. 

4.  Fig.  116  shows  a  pattern  consisting  of  hexagons  and  equi- 
lateral triangles  combined. 


Fig.  116. 


Fig.  117. 


5.  Regular  octagons  alone  are  not  sufficient;  for  the  angle  of 
the  octagon  is  135°.  and  twice  135°  is  only  270°,  and  there  remains 
an  angle  of  90°  to  be  filled.  This  can  be  done  by  the  right  angle 
of  a  square  (Fig. 


132 


GEOMETRY    FOR    BEGINNERS. 


[§ 


Fig.  118. 


6.   The  angle  of  a  regular  decagon  is   144°.      Two  decagons 

placed  with  one  side  common 
would  leave  an  angle  of  360°  —  2 
X  144  =  72°  to  be  filled.  Since 
no  regular  polygon  has  an  angle 
of  72°,  it  follows  that  we  cannot 
pave  or  quilt  with  regular  deca- 
gons, either  alone  or  combined 
with  other  regular  polygons. 

7.  The  regular  dodecagon  may 
be  combined  with  the  equilateral 
triangle ;  for  the  angle  of  the  do- 
decagon is  150°,  twice  150°  is  300°,  and  there  remains  60°,  which 
is  just  equal  to  the  angle  of  the  equilateral  triangle  (Fig.  118) . 

8.  Thus  far  we  have  spoken  only  of  regu- 
lar polygons.  It  is  obvious  that  suitable  com- 
binations may  also  be  made  with  polygons 
which  are  not  regular.  Fig.  119  gives  an 
example.  What  is  the  name  of  the  polygon? 

Exercise.  —  Show  that  a  square,  a  hexagon,  and  a 
dodecagon  will  fill  up  the  space  about  a  point ;  and 
make  a  pattern  of  these  polygons. 

III.  —  Symmetrical  Figures. 

§  123.    If  the  polygon  ABCD  is  made  to  revolve  about  AD 
through  half  a  revolution,  the  figure  AEFD,  thereby  obtained,  is 

said  to  be  SYMMETRICAL  with  respect 
to  ABCD,  and  AD  is  called  the 
LINE  OF  SYMMETRY. 

Two  symmetrical  plane  figures  are 
always  equal,  but  their  equal  parts 
(sides  and  angles)  follow  in  opposite 
order  with  reference  to  the  line  of 
symmetry. 


§    I  23.]  CHAPTER  VI.  —  POLYGONS.  133 

From  what  precedes  it  is  obvious  that  in  two  symmetrical  plane 
figures,  — 

I.    Two  symmetrically-situated  sides  are  equal. 
II.    Two  symmetrically-situated  angles  are  equal. 

III.  The  line  joining  two  symmetrically-situated  points  is  per- 
pendicular to  the  line  of  symmetry  and  bisected  by  it  (for  example, 
the  lines  BE,  MN,  CF,  etc.,  Fig.  126}. 

By  means  of  III.  we  can  easily  construct  upon  one  side  of  a 
polygon,  as  a  line  of  symmetry,  a  symmetrical  polygon.  Explain 
how. 

If  we  regard  ABCDFE  as  one  polygon,  then  AD  is  a  line 
of  symmetry  dividing  it  into  two  symmetrical  polygons. 

A  line  may  divide  a  figure  into  parts  which  are  equal  and  yet 
not  symmetrical.  This  is  the  case,  for  example,  with  the  diagonal 
of  the  parallelogram  A  BCD  (Fig.  94).  It  divides  the  figure 
into  two  equal  triangles,  but  these  triangles  are  not  so  placed  with 
respect  to  the  diagonal,  that  they  would  coincide  if  folded  over  the 
diagonal  as  an  edge. 

A  figure  may  have  more  than  one  line  of  symmetry.  Thus,  in 
Fig.  1 20,  besides  AD,  the  line  MN  is  also  a  line  of  symmetry. 

Exercises.  —  1.  Point  out  in  Fig.  120  symmetrically-situated  lines  and 
angles.  « 

2»  Draw  a  parallelogram,  and  upon  one  side  construct  a  symmetrical 
parallelogram. 

3*    Construct  upon  one  side  of  a  hexagon  a  symmetrical  hexagon. 

4.  Construct  two  symmetrical  figures  with  reentrant  angles. 

5.  Draw  a  line  of  symmetry  in  an  isosceles  triangle. 

6.  Draw  all  the  lines  of  symmetry  in  an  equilateral  triangle. 
7«    Draw  all  the  lines  of  symmetry  in  a  square. 

8.  Draw  all  the  lines  of  symmetry  in  a  rectangle. 

9.  Make  a  trapezoid  such  that  you  can  draw  in  it  a  line  of  symmetry. 

10.  Give  instances  of  figures  which,  placed  together  with  a  side  common, 
will  make  a  symmetrical  figure.     What  is  the  line  of  symmetry? 

11.  Draw  all  the  lines  of  symmetry  in  a  regular  hexagon.     How  many  are 
there  ? 


134  GEOMETRY    FOR    BEGINNERS. 

REVIEW   OF   CHAPTER  VI. 
SYNOPSIS. 

I 

1.  A  polygon  is  a  plane  figure  bounded  by  straight  lines. 

2.  Polygons  receive  different  names  according  to  the  number  of  their  sides. 

3.  The  sum  of  the  angles  of  a  polygon  =  180°  X  the  number  of  sides  less 
two. 

4.  Two  polygons  are  equal,  if  they  can  be  divided  into  the  same  number 
of  triangles,  similar  each  to  each,  and  similarly  placed. 

5*    A  polygon  is  equilateral,  if  its  sides  are  equal ;   equiangular,  if  its  angles 
are  equal ;   regular,  if  both  sides  and  angles  are  equal. 

6,  Each  angle  of  a  regular  polygon  may  be  found  by  dividing  360°  by  the 
number  of  sides,  and  then  subtracting  the  quotient  from  180°. 

7.  In  every  regular  polygon  there  is  a  point  called  the  centre,  equally  dis- 
tant from  its  corners  and  also  from  its  sides. 

8»   The  greater  radius  of  a  regular  polygon  is  the  distance  from  the  centre 
to  a  corner  ;   the  less  radius  the  distance  from  the  centre  to  a  side ;   the 
angle  at  the  centre  the  angle  included  between  two  greater  radii. 
9»    Paving  and  quilting  furnish  examples  of  the  practical   use   of  regular 
polygons. 

10»  Two  plane  figures  are  symmetrical,  if  either  of  them,  when  revolved 
about  a  certain  line  as  an  axis  through  half  a  revolution,  can  be  made 
to  fall  upon  and  coincide  with  the  other.  The  line  is  called  the  line  of 
symmetry. 

11.  There  are  figures  in  which  several  lines  of  symmetry  may  be  drawn,  each 
dividing  the  figure  into  two  symmetrical  figures. 

EXERCISES. 

1.  Make  a  hexagon  having  four  right  angles. 

2.  Make  an  octagon  having  only  right  angles  or  their  multiples. 

3.  Make  three  polygons  with  reentrant  angles. 

4.  Divide  a  hexagon  into  other  figures  in  several  different  ways. 

5.  Find  the  sum  of  the  angles  of  a  polygon  with  24  sides. 

6.  Find  each  angle  of  a  regular  icosagon. 

7.  Eight  angles  of  a  decagon  are  equal,  and  each  of  the  other  two  is  twice 
one  of  the  former  ;   find  all  the  angles. 


CHAPTER   VI. REVIEW.  135 

8.  In   a   dodecagon  each  angle  taken  in  order  is   i°  20'  more  than  the 
preceding  ;    find  all  the  angles. 

9.  Find  the   angles  of  a  pentagon,  if  they  are  to  each  other  as  the  num- 
bers I,  2,  3,  4,  5. 

10.  Draw  a  heptagon ;    then  construct  another  equal  to  it. 

11.  Construct  a  regular  decagon. 

12.  Make  a  five-rayed  star,  and  prove  that  the  sum  of  the  five  acute  angles  is 
equal  to  1 80°. 

Hint.  —  The  star  is  made  by   prolonging  the  sides  of  a  regular  pentagon  till 
they  meet,  and  then  erasing  the  sides  of  the  pentagon. 

13.  Make  a  seven-rayed  star,  and  find  the  sum  of  the  seven  acute  angles. 
14:.    Can  you  make  a  pattern  of  squares,  pentagons,  and  icosagons? 

15.  Can  you  make  a  pattern  of  pentagons  and  decagons? 

16.  Construct  a  triangle,  and  upon  its  longest  side  a  symmetrical  triangle. 

17.  Construct  a  rhombus  ;    then  construct  upon  one  of  its  sides  a  symmetrical 
rhombus. 

18.  Draw  a  polygon  with  two  reentrant  angles,  and  then  construct  a  sym- 
metrical polygon. 

19.  Make  a  right  triangle  in  which  a  line  of  symmetry  can  be  drawn. 

20.  How  many  lines  of  symmetry  can  be  drawn  in  a  regular  polygon  of  any 
number  (n)  of  sides  ? 


136  GEOMETRY   FOR   BEGINNERS.  [§   124. 


CHAPTER  VII. 
AREAS. 

CONTENTS.—  I.  Units  of  Area  (§§  124, 125).  II.  The  Areas  of  Polygons  (§§  126- 
135).  III.  Practical  Exercises  and  Applications  (§§  136-141).  IV.  Theorem  of 
Pythagoras  (§§  142,  143).  V.  Transformation  of  Figures  (§§  144-150).  VI.  Par- 
tition of  Figures  (§§  151-155). 

L— Units  of  Area. 

§  124.  Surfaces,  like  lines,  are  measured  by  choosing  a  unit, 
and  then  finding  how  often  this  unit  is  contained  in  the  surface 
which  we  wish  to  measure. 

The  unit  chosen  for  this  purpose  must  be  a  magnitude  of  the 
same  kind  as  that  which  is  to  be  measured ;  in  other  words,  it 
must  be  a  surface.  The  most  convenient  units  to  employ  are 
squares  whose  sides  are  equal  to  the  iinits  of  length. 

Chief  English  units  :  the  SQUARE  INCH  (sq.  in.),  the  SQUARE  FOOT 
(sq.  ft.),  the  SQUARE  YARD  (sq.  yd.),  and  the  SQUARE  MILE. 
30^  sq.  yds.  =  i  SQUARE  ROD,  and  160  sq.  rods  =  i  ACRE. 

The  metric  units  (with  their  abbreviations1  and  relative  values) 
are  as  follows  :  — 

SQUARE  KILOMETER  (qkm)    =  1,000,000  square  meters. 

SQUARE  HECTOMETER  (qhm)  =  10,000 

SQUARE  DEKAMETER  (qdkm)  =  100 

SQUARE  METER  (qm)             =  i 

SQUARE  DECIMETER  (qdm).  =  o.oi 

SQUARE  CENTIMETER  (qcm)  =  o.oooi 

SQUARE  MILLIMETER  (qmm)  =  o.oooooi 

i  Formed  by  prefixing  q,  the  first  letter  of  the  Latin  word  quadra  (a  square)  to 
the  corresponding  linear  abbreviations. 


I25-] 


CHAPTER    VII.  — -  AREAS. 


137 


D 


The  square  dekameter  is  usually  called  an  AR  (a),  and  the 
square  hectometer  a  HECTAR  (ha).  They  are  employed  chiefly 
in  measuring  land. 

NOTE.  —  iba  =  2.471  acres  ;   iim  =  10.764  sq.  ft. ;  iidm  =  15.5  sq.  in.,  very  nearly. 

§  125.  In  the  table  of  metric  units  above  given,  it  will  be 
observed  that  each  unit  is  100  times  greater  than  the  unit  next 
following.  The  reason  of  this  lies  in  the  fact  that,  of  the  corre- 
sponding linear  units,  one  is  10  times  greater  than  the  other  (see 
page  41). 

Suppose  that  A  B  (Fig.  121)  repre- 
sents a  length  of  one  meter.  Con- 
struct upon  AB  the  square  A  B  C D ; 
then  this  square  will  represent  one 
square  meter.  Divide  A  B  and  A  C 
each  into  ten  equal  parts,  and  through 
all  the  points  of  division  draw  lines 
parallel  to  the  adjacent  sides.  The 
parallels  through  the  points  of  division 

of  AB  divide  the  square  into  10  equal      A  Fia.  I2I  & 

rectangulars,  each   iodm  long  and  idm 

wide  ;  the  other  set  of  parallels  subdivides  each  rectangle  into 
10  equal  squares,  each  equal  to  one  square  decimeter ;  there- 
fore, one  square  meter  must  contain"  exactly  10  X  10=100  square 
decimeters. 

The  same  reasoning  will  show  that  one  square  decimeter  =  10 
X  10  =  100  square  centimeters,  etc. ;  and  if  the  ratio  of  the  two 
linear  units  were  any  other  number  but  10,  it  may  be  shown  in 
the  same  way  that  the  ratio  of  the  corresponding  units  of  surface 
would  be  found  by  multiplying  that  number  by  itself;  in  other 
words,  by  squaring  it. 

The  ratio  of  two  units  of  surface  is  always  the  square  of  the 
ratio  of  the  corresponding  units  of  length. 


138  GEOMETRY    FOR    BEGINNERS.  [§    126. 

Exercises.  —  1.    What  is  the  ratio  of  a  square  foot  to  a  square  inch?  a 
square  foot  to  a  square  yard?  a  square  mile  to  a  square  foot? 

2.  How  many  square  feet  are  there  in  one  acre  ? 

3.  How  many  acres  are  there  in  one  square  mile  ? 

4.  What  is  the  ratio  of  an  ar  to  a  square  meter  ?  an  ar  to  a  hectar?  a 
square  centimeter  to  a  hectar?  a  square  hectometer  to  a  hectar? 

5*    Reduce  to  square  meters  the  following:  4<ikm;    I5ha;   9-87a;   o.o84ha; 
97,6751™. 

6.  Reduce  to  hectars  the  following:  63<ikm;   39. 2a;   36ooim;   $6,4-75qAm> 
4,ooo,oooimm. 

7.  Subtract  ia  from  iha,  and  give  the  answer  in  square  meters. 

8.  How  many  house-lots,  each  containing  2a,  are  there  in  a  field  which 
contains  8ha? 

9.  A  man  bought  3ha  of  land  at  $200  per  hectar,  and  sold  it  for  $2.50  per 
ar.     Did  he  gain  or  lose,  and  how  much? 

10.  Which  is  the  greater  farm,  200  hectars  or  500  acres? 

11.  Divide  6ha  into  64  equal  lots  ;   how  many  square  meters  in  each  lot? 

12.  What  is  the  difference  between  3  square  meters  and  3  meters  square? 


II.— Areas  of  Polygons. 

§  126.  Definition.  —  The  number  of  times  a  unit  of  surface 
is  contained  in  any  surface,  followed  by  the  name  of  the  unif,  is 
called  the  AREA  of  the  surface. 

NOTE.  —  Hence  units  of  surface  are  often  called  units  of  area,  as  at  the  begin- 
ning of  this  chapter. 

In  order  to  find  the  area  of  a  surface,  it  might  at  first  thought 
seem  necessary  to  apply  a  unit  of  area  to  the  surface  over  and 
over  again  as  many  times  as  possible,  just  as  we  measure  a  line 
directly  by  using  a  yard-stick  or  a  meter-rule.  But  this  method 
would  be  very  tedious,  and  in  many  cases  (for  instance,  a  pond, 
swamp,  forest,  etc.)  utterly  impracticable. 

Fortunately,  however,  Geometry  supplies  us  with  indirect  meth- 
ods of  measurement  which  are  applicable  to  all  surfaces.  It  teaches 
that  the  value  of  areas  depends  upon  the  lengths  of  certain  lines ; 
whence  it  follows  that  areas  can  be  found  by  performing  certain 


§  127.]  CHAPTER  VII. — AREAS.  139 

simple  operations  upon  the  numbers  which  express  the  lengths  of 
these  lines. 

In  explaining  these  indirect  methods  we  begin  with  the  square 
and  the  rectangle,  because  they  can  be  readily  decomposed  into 
square  meters,  or  into  squares  larger  or  smaller  than  the  square 
meter. 

§  127.  The  SQUARE.  If  the  side  of  a  square  contains  a  whole 
number  of  meters,  as  six,  it  is  easy  to  see  that  (by  proceeding 
exactly  as  in  §  125)  we  can  first  divide  the  square  into  six  equal 
rectangles,  then  each  rectangle  into  six  square  meters;  so  that 
the  square  will  contain  in  all  6  X  6  =  36  square  meters. 

The  same  mode  of  reasoning  may  be  used  when  the  side  con- 
tains meters  and  subdivisions  of  a  meter.  Let,  for  instance,  the 
side  =  6.35ra;  then,  in  decomposing  the  square  into  smaller  squares, 
we  may  take  as  the  unit  of  length  one  centimeter.  In  place  of  six 
rectangles,  there  will  be  635  ;  and  in  place  of  6  X  6  =  36  square 
meters,  there  will  be  635  X  635  =  403,225  square  centimeters  = 
40.3225  square  meters,  —  a  result  obtained  simply  by  multiplying 
6.35  by  itself. 

In  short,  the  number  of  units  of  area  in  a  square  is  always  found 
by  multiplying  by  itself  the  number  of  the  corresponding  units  of 
length  in  one  of  its  sides. 

Hence,  when  we  multiply  a  number  by  itself,  we  are  said  to  square 
it ;  and  it  is  usual  to  express  the  preceding  rule  more  briefly  thus  : 
The  area  of  a  square  is  equal  to  the  square  of  one  of  its  sides. 

Conversely,  if  the  area  is  known,  in  order  to  find  the  value  of 
one  side,  we  must  find  a  number  which,  multiplied  by  itself,  will 
give  the  numerical  value  of  the  area ;  in  other  words,  we  must 
extract  the  square  root  of  the  given  area. 

For  example  :  if  the  area  of  a  square  =  8.1225  square  meters, 
me  side  =  V8.I225  =  2.85  meters. 

Remark.  —  For  the  sake  of  convenience  we  use  the  expression 


140 


GEOMETRY    FOR    BEGINNERS. 


[§    128. 


"the  square  of  AB"  and  we  write  AB*-  These  expressions  do 
not  mean  that  the  line  AB  is  multiplied  by  itself,  but  that  the 
number  of  linear  units  in  the  line  is  multiplied  by  itself,  the  result 
being  the  number  of  units  of  area  in  the  square  constructed  upon 
AB  as  a  side. 

§  128.   The  RECTANGLE.     A  rectangle  can  be  decomposed  into 
units  of  area  in  the  same  way  as  a  square,  the  only  difference  being 

that  the  number  of  units  of  length  in 
its  adjacent  sides  is  not  the  same. 

Thus,  the  rectangle  ABCD  (Fig. 
122),  the  sides  of  which  are  A£=  ym, 
AC=4m,  contains  7x4=28  square 
meters. 

The  side  AB  is  the  base  of  the  rect- 


D 


Fig.  122. 


angle,  and  the  side  AC  is  the  altitude  (§  104) ;  so  that  the  number 
of  units  of  area  in  a  rectangle  is  found  by  multiplying  the  number 
of  linear  units  in  the  base  by  the  number  of  linear  units  in  the 
altitude. 

Or,  more  briefly :  The  area  of  a  rectangle  is  equal  to  the  pro- 
duct of  its  base  by  its  altitude. 

If  the  area  and  base  are  known,  how  can  the  altitude  be 
found?  If  the  area  and  altitude  are  known,  how  can  the  base 
be  found? 

Remark.  —  In  practice  we  must  be  careful  to  express  both 
base  and  altitude  in  terms  of  the  same  linear  unit ;  the  area  will 
then  be  expressed  in  terms  of  the  corresponding  unit  of  area. 

For  example  :  the  base  of  a  rectangle  =  i2m,  the  altitude  =  8ocm ; 
find  the  area.  Here  we  must  not  multiply  12  by  80 ;  but,  if  we 
choose  the  meter  as  the  linear  unit,  we  must  multiply  12  by  0.8, 
because  8ocm  =  0.8™.  Answer,  9.6  square  meters. 

If  we  take  the  centimeter  as  the  linear  unit,  what  will  the 
answer  be? 


§ 


CHAPTER   VII. AREAS. 


141 


Fig.  123. 


§  129.  The  PARALLELOGRAM.  It  is  easy  to  transform  a  paral- 
lelogram into  an  equivalent  rectangle. 

In  the  parallelogram  ABCD  (Fig.  123)  erect  perpendiculars 
at  the  points  A  and  B  of  the  base ; 
they  cut,  one  of  them  the  opposite 
side,  the  other  the  opposite  side  pro- 
longed, in  the  points  F  and  E ;  and 
the  figure  ABEF  is  a  rectangle  having 
the  same  base  and  altitude  as  the 
parallelogram. 

The  right  triangles  AC  E  and  B  DP  are  equal  (why?)  ;  hence, 
whichever  of  these  triangles  we  add  to  the  quadrilateral  ABCF, 
we  shall  obtain  the  same  area.  If  we  add  AAC£,  we  obtain  the 
rectangle  ABEF]  and  if  we  add  ABFD,  we  obtain  the  paral- 
lelogram ABCD:  therefore,  the  parallelogram  ABCD  is  equal 
to  the  rectangle  ABEF. 

Theorem.  — A  parallelogram  is  equivalent  to  a  rectangle  having 
the  same  base  and  altitude. 

NOTE.  —  This  transformation  maybe  easily  shown  to  the  eye  by  cutting  from 
stiff  paper  the  trapezoid  A  CBF  {Fig.  123),  and  the  right  triangle  A  E  C,  and  then 
placing  the  triangle  in  the  two  positions  shown  in  the  figure. 

Corollaries.  —  i.  By  combining  this  theorem  with  §  128,  we 
see  that  the  area  of  a  parallelogram  is  equal  to  the  product  of  its 
base  by  its  altitude. 

2.   All  parallelograms  having  equal  bases  and 
equal  altitudes  are  equivalent. 

Is  the  converse  of  Corollary  2  true  ?     Draw 
a  diagram  to  illustrate. 

Construct  several  equivalent  parallelograms 
differing  in  shape. 

Remark.  —  In    a   rhombus   the    diagonals 
bisect  each  other  at  right  angles  (§  104). 

Construct  about  the  rhombus  ABCD  (Fig. 
124)  a  rectangle  PQMN,  by  drawing  through  the  corners  of  the 


Fig.  124. 


142 


GEOMETRY    FOR    BEGINNERS. 


[§  130. 


rhombus  lines  parallel  to  its  diagonals.  It  is  easy  to  show  that  this 
rectangle  is  composed  of  eight  equal -right  triangles,  of  which  the 
rhombus  contains  four  (how?).  Now,  since  the  base  and  altitude 
of  the  rectangle  are  equal  to  the  two  diagonals  of  the  rhombus,  its 
area  is  equal  to  the  product  of  these  diagonals.  Hence,  the  area 
of  the  rhombus  is  equal  to  half  the  product  of  its  diagonals. 

§  130.  The  TRIANGLE.  A  triangle  ABC  (Fig.  125)  may  be 
regarded  as  half  a  parallelogram  having  an  equal  base  and  equal 
altitude  (§  101).  We  have  only  to  draw  through  two  corners,  B 
and  C,  of  the  triangle  lines  parallel  to  the  op- 
posite sides  ;  then,  since  &ABC=%  ABDC, 
and  ABDC  =  AB  X  CE,  therefore  A  ABC 
=  %A£xC£.  That  is,  - 

The  area  of  a  triangle  is  equal  to  half  the 
product  of  its  base  by  its  altitude. 

Fi  Corollary.  —  All   triangles   having  equal 

bases  and  equal  altitudes  are  equivalent. 

NOTE.  —  In  finding  the  area  of  a  triangle,  we  may  either  multiply  the  base  by  the 
altitude,  and  then  divide  by  2,  or  we  may  multiply  half  the  base  by  the  altitude,  or 
we  may  multiply  half  the  altitude  by  the  base.  For  example  :  if  the  base  =  14"',  and 


the  altitude  =  8m,  the  area  = 


14X8 


14  X  -  =  56  square  meters. 


§  131.  The  TRAPEZOID.  A  trapezoid  ABCD  (Fig.  126)  is 
divided  by  the  diagonal  BC  into  two  triangles,  ABC  and 
BCD,  whose  bases,  AB  and  CD,  are  the 
parallel  sides  of  the  trapezoid,  and  whose 
common  altitude  CE  is  the  altitude  of  the 
trapezoid. 

Since  A  ABC  =  £  AB  X  CE, 
and  A  BCD  =  }  CD  X  CE; 
therefore,  the  trapezoid  ABCD  =  %(AB+  CD}  X  CE. 
That  is,  the  area  of  a  trapezoid  is  equal  to  half  the  product  of 
the  sum  of  the  parallel  sides  by  the  altitude. 


§ 


CHAPTER   VII. AREAS. 


143 


§  132.  Any  QUADRILATERAL.  To  find  the  area  of  any  quadri- 
lateral, divide  it  by  a  diagonal  into  two  triangles,  find  the  areas  cf 
these  triangles,  and  add  the  results  together. 

§  133.  A  REGULAR  POLYGON.  We  know  (§  119,  Corollary  i) 
that  a  regular  polygon  ABCDEF  (Fig.  127)  is  composed  of  as 
many  equal  isosceles  triangles  as  there 
are  sides.  Therefore  its  area  may  be 
found  by  adding  together  the  areas  of 
these  triangles.  The  altitude  of  each 
triangle  is  equal  to  the  less  radius  of 
the  polygon,  so  that  the  area  of  each 
triangle  is  half  the  product  of  its  base 
by  the  less  radius.  Therefore  the  area 
of  the  polygon  is  half  the  product  of  the 
sum  of  all  the  bases  (that  is  to  say,  the 
perimeter  of  the  polygon)  by  the  common  altitude  (that  is  to  say, 
the  less  radius  of  the  polygon) . 

The  area  of  a  regular  polygon  is  half  the  product  of  its  perimeter 
by  its  less  radius. 

§  134.  Any  POLYGON.  The  area  of  any  polygon  may  be  found 
by  dividing  it  into  triangles.  Fig.  128  shows  two  ways  of  doing 
this.  In  either  case,  if  we 
compute  the  area  of  each 
triangle  separately,  and 
then  add  the  results,  we 

shall  find  the  area  of  the  \  \    /  /  \    / 

polygon. 

In  each  of  these  cases 


Fig.  I2J. 


Fig.  128. 


how  many  lines  require  to  be  measured  directly  in  order  to  com- 
pute the  area? 

But  it  is  generally  easier  to  find  the  area  of  an  irregularly-shaped 


144 


GEOMETRY    FOR    BEGINNERS. 


[§  134- 


polygon  (as  a  field,  farm,  etc.)  by  division  into  right  triangles  and 

trapezoids,  as  shown  in  Fig. 
129.  Join  the  two  most  dis- 
tant corners  of  the  polygon 
by  a  straight  line,  and  let  fall 
from  the-  other  corners  per- 
pendiculars to  this  line  taken 
as  a  common  base.  The 
polygon  is  thus  divided  into 
right  triangles  and  trapezoids, 
the  areas  of  which  are  then 
computed  separately,  and 

their  sum  is  obviously  equal  to  the  area  of  the  polygon. 

For  example  :  let  (Fig.  129)  Bb  =  6.8m  ;  Cc  =  io.6m  ;  Dd  = 
io.im;  Ff  =  8.3m;  Gg  =  6.2m  ;  Nh  =  9.2™  :  also  let  Ab  =  5.6™; 
th  =  2.6m;  /^  =  4.2m;  ^=4.6m;  gf  =  3m  ;  fd  =  2.8m;  dE 


The  work  of  computation  may  be  arranged  as  follows  :  — 


FACTORS. 

PARTS  OF  THE 
POLYGON. 

BASES,   OR  SUMS 

PRODUCTS. 

OF   PARALLEL   SIDES. 

ALTITUDES. 

&ABb 

Bb=    6.8 

Ab=    5.6 

38.08 

Trap.  Bbc  C 

Bb  +   Cc=  17.4 

bc=    6.8 

118.32 

Trap.  CcDd 

Cc  -}-  Dd=  20.7 

c  d=  10.4 

215.28 

l\DdE 

Dd—  10.1 

<*£=    5-8 

58.58 

&FfE 

Ff=    8.3 

/£=    8.6 

7I-38 

Trap.  Ffg  G 

*f+  Gg  =  14.5 

/^r=  s 

43-50 

Trap.  GghH 

Gg+Hh~-  15.4 

gh=    8.8 

I35-52 

kAhH 

Ah-    8.2 

^>^=    9.2 

75-44 

2)756.10 

Polygon  ABCDEFGH  = 

378.o5qm 

§  135-3 


CHAPTER   VII. AREAS. 


145 


NOTE  i.  —  In  the  preceding  table,  in  place  of  dividing  each  product  by  2,  we  first 
add  the  products,  and  then  divide  their  sum  by  2. 

NOTE  2.  —  In  case  the  interior  of  the  polygon  is  inaccessible,  Fig.  ijo  illustrates 
one  way  of  meeting  the  difficulty.  Construct  a  rectangle  which  shall  completely 
enclose  the  polygon,  and,  from  all  the  corners  of  the  polygon  which  are  not  in  the 


Fig.  130. 

sides  of  the  rectangle,  let  fall  perpendiculars  to  the  sides  of  the  rectangle.  In  this 
way  a  number  of  right  triangles  and  trapezoids  are  formed,  the  areas  of  which  can 
be  found  in  the  usual  way ;  and  it  is  evident  that  if  the  sum  of  their  areas  is  sub- 
tracted from  the  area  of  the  rectangle,  the  remainder  will  be  equal  to  the  area  of  the 
polygon. 

§  135.  It  will  now  be  clear  in  what  way  Geometry  supplies  us 
with  indirect  methods  of  finding  areas.  In  the  first  place,  it  teaches 
that  the  area  of  a  rectangle  (the  square  is  simply  a  rectangle  with 
equal  sides)  depends  entirely  on  the  lengths  of  its  two  sides,  and 
can  therefore  be  found  as  soon  as  the  lengths  of  these  sides  are 
known ;  it  then  shows  that  any  parallelogram  can  be  transformed 
into  an  equivalent  rectangle,  that  a  triangle  is  half  a  parallelogram 
of  the  same  base  and  altitude,  and  that  a  trapezoid  consists  of  two 
triangles  of  equal  altitude  ;  lastly,  it  shows  that  any  polygon  what- 
soever may  be  decomposed  into  triangles  and  trapezoids. 

Remark.  —  In  some  cases  the  rule  for  finding  an  area  may  be 
very  briefly  expressed  by  a  formula  in  which  letters  are  used  to 
represent  the  values  of  the  various  quantities. 

Let  6"  denote  in  all  cases  the  area,  a  the  side  of  a  square,  a  and 
b  the  adjacent  sides  of  a  rectangle,  a  and  h  the  base  and  altitude 


146  GEOMETRY    FOR    BEGINNERS.  [§   136. 

of  a  parallelogram  or  triangle,  a  and  c  the  parallel  sides,  h  the 
altitude  of  a  trapezoid,  p  and  r  the  perimeter  and  less  radius  of 
a  regular  polygon.  Then,  — 


For  the  square, 
For  the  rectangle, 
For  the  parallelogram, 
For  the  triangle, 
For  the  trapezoid, 
For  the  regular  polygon, 

S  =  a2, 
S  =  a  X  b, 
S  =  a  x  h, 
S  =  \  a  X  h, 
S  =  Ka  +  c)  X  h, 
S=fpxr, 

[3-1 
[4-] 

T5-] 
[6.] 

L7-] 
[8.] 

III.  —  Practical  Exercises  and  Applications. 

NOTE.  —  In  solving  most  of  the  exercises,  it  is  well  to  draw  (free-hand)  a  figure 
representing  as  nearly  as  possible  the  true  shape  of  the  polygon. 

§  136.   THE  SQUARE  (S  =  a*). 

In  a  square,  given,  — 

1.  The  side  =  23™;   find  the  area. 

2.  The  side  —  35™;   find  the  area. 

3.  The  side  =  4|ft;  find  the  area. 

4.  The  side  =  5fyds;  find  the  area. 
6.   The  side  =  6.25™;   find  the  area. 

6.  The  side  =  48.72™;   find  the  area. 

7.  The  area  =6259™;   find  the  side. 

8.  The  area  =  515. 29<im;   find  the  side. 

9.  The  area  =  63.2025<3m;   find  the  side. 

10.  The  area  =  54.76*;   find  the  side  in  meters. 

11.  The  area  =  i6.32i6ha;   find  the  side  in  meters. 

12.  The  area  =  78.8544^;   find  the  side  in  dekameters. 

13.  One  side  of  a  square  garden  measures  37.6™;   find  its  area. 

14.  A  square  park  contains  64  acres.     How  many  trees,  2oyds  apart,  can 
be  set  out  around  it  ? 

15.  A  square  park  contains  64ha.     How  many  trees,  2Om  apart,  can  be  set 
out  around  it  ? 

16.  The  three  dimensions  of  a  room  are  each  equal  to  6.4™.     What  will  it 
cost  to  paper  its  walls,  if  the  price  is  $0.15  per  square  meter  ? 

17.  Find  the  cost  of  looo  panes  of  glass,  each  o.48m  square,  if  the  glass 
is  worth  $1.40  per  square  meter. 


§    1 3  7.]  CHAPTER   VII. AREAS.  147 

18.  Area  of  the  State  of  New  York  =  47,156  square  miles.     Find  the  side 
of  a  square  having  the  same  area. 

19.  Construct  a  square,  and  then  find  its  area. 

§  137.   THE  RECTANGLE  (S  =  a  X  i>}. 

In  a  rectangle,  given,  — 

20.  Length  =  6om,  breadth  =  24™;   find  the  area. 

21.  Length  =  2m,  breadth  =  8ocm;   find  the  area. 

22.  Length  =  8.5™,  breadth  =  40™;   find  the  area. 

23.  Length  -  92dkm,  breadth  =  i6.37dkm;   nnc1-  tne  area  in  hectars. 

24.  Length  =  28fft,  breadth  =  i"jlft;   find  the  area  in  sq.  yds. 

25.  Length  =  627?^,  breadth  =  435^'^;   find  the  area  in  acres. 

26.  Area  =  3601™,  length  =  30™;   find  the  breadth. 

27.  Area  =  5.6%  length  =  i6m;   find  the  breadth. 

28.  Area .=  50^,  length  =  fkm;   find  the  breadth. 

29.  Area  =  6  acres,  breadth  =  44^ ;   find  the  length. 

30.  Area  =  767.24<im,  breadth  =  24.5™;    find  the  length. 

31.  Area  =  7.2ha,  breadth  =  o.25kil;   find  the  length. 

32.  Find  the  area  of  the  surface  of  a  table  1.58™  long,  i.2m  wide. 

33.  How  many  ars  in  a  field  87.5™  by  45. 8m? 

34.  Find  the  area  of  a  field  24Om  long  and  T\  as  wide. 

35.  The  perimeter  of  a  rectangle  =  24™,  the  length  =  9.2™;  find  the  breadth 
and  the  area. 

36.  If  the  area  of  a  cellar  48™  long  is  15001™,  what  is  its  breadth? 

37.  The  perimeter  of  a  rectangle  =  86.2m,  the  breadth  =  12. 6m;  find  the 
length  and  the  area. 

38.  The  area  of  a  rectangle  =  I44qm,  the  breadth  =  6m;  find  the  length  and 
the  perimeter. 

39.  The  area  of  a  rectangle  =  S,  the  length  —  a  ;   find  the  breadth  and  the 
perimeter. 

40.  Construct  a  rectangle,  and  then  find  its  area. 

41.  Find  the  area  of  the  floor  of  the  room. 

42.  The  perimeter  of  a  rectangle  =  72™,  and  the  length  is  twice  the  breadth; 
find  length,  breadth,  and  area. 

43.  If  the  perimeter  =  96"*,  and  the  length  is  to  the  breadth  as  7:5,  find 
length,  breadth,  and  area. 

44.  If  the  perimeter  =  26™,  and  the  length  is  2-5m  more  than  the  breadth, 
find  length,  breadth,  and  area. 


148  GEOMETRY   FOR   BEGINNERS.  [§   137. 

45.  How  much  larger  is  a  reotangle  if  the  length  is  doubled  ?  if  the  breadth 
is  doubled  ?   if  both  are  doubled  ? 

46.  The  sides  of  a  rectangle  are  42™  and  6om;   find  the  side  of  a  square 
equivalent  to  the  rectangle. 

47.  If  the  side  of  a  square  —  I2m,  find  the  sides  of  an  equivalent  rectangle. 
Is  there  any  thing  peculiar  about  this  exercise  ? 

48.  A  square  and  a  rectangle  have  the  same  perimeter,  6om,  and  the  rect- 
angle is  twice  as  long  as  it  is  wide.     Find  their  areas. 

49.  Construct  a  rectangle   i6dm  long,  4dm  wide  ;   then  another,  I5dm,  5dm 
wide  ;   then  another,  I4dm,  6dm  wide  ;   and  so  on,  until  you  reach  a  square. 
Then  compare  perimeters  and  areas.     Which  has  the  greatest  area  ? 

50.  A  chain  8om  long  is  made  to  enclose  a  rectangle,  one  side  being  25™. 
How  much  more  area  would  it  enclose  if  the  figure  were  a  square  ? 

51.  A  and  B  exchange  fields  of  equally  good  soil.     A's  field  is  rectangular 
in  shape,  4  times  as  long  as  wide,  and  with  a  perimeter  =  iooom;   B's  field  is 
in  the  shape  of  a  square,  the  perimeter  being  8oom.     Which  has  the  best  of 
the  bargain  ? 

52.  From  a  rectangular  field  528m  by  240™  a  piece  6om  square  is  sold. 
How  much  is  left  ? 

Find  the  number  of  linear  meters  of  carpeting  required  for  floors  of  the 
following  dimensions :  — 

53.  Floor,  8m  by  6.6m;   carpet,  im  wide. 

54.  Floor,  9.5™  by  7.7™;  carpet,  8ocm  wide. 

55.  Floor,  I2m  by  8.4m;   carpet,  i.2m  wide. 

56.  Floor,  24ft  by  i6ft  8in;   carpet,  3O5n  wide. 

Find  the  cost  of  carpeting  rooms,  their  dimensions  and  the  width  and  price 
of  carpeting  being  as  follows  :  — 

57.  Floor,  iom  by  6.5™;   carpet,  gocm  wide,  at  $1.25  per  meter. 

58.  Floor,  11.4™  by  5m;   carpet,  1.4™  wide,  at  $2  per  meter. 

59.  Floor,  8m  by  7.37™;   carpet,  6ocm  wide,  at  $0.75  per  meter. 

60.  Floor,  i8^ft  by  14^;   carpet,  |yd  wide,  at  $1.60  per  yard. 

61.  How  much  oil-cloth  will  it  take  to  line  the  inside  of  a  box  2.6m  by  i.2m 
by  o.8m  ? 

62.  How  many  tiles  8in  square  will  cover  a  floor  36ft  by  2Oft? 

63.  What  will  it  cost  to  paint  the  floor  and  walls  of  a  room  8m  long,  6m 
wide,  3m  high,  at  one  cent  per  square  centimeter  ? 

64.  Find  the  cost  of  glazing  12  windows,  each  6ft  6in  by  3ft  4in,  at  $0.50 
per  square  foot  ? 

65.  A  room  is  iom  long,  6m  wide,  and  3-6m  high.  There  are  four  windows, 


§   138.]  CHAPTER   VII. AREAS.  149 

each  2.2™  by  1.2™;  two  doors,  each  3™  by  i.8m;  and  the  base-board  is  9Ocm 
high.  Required  (zV)  the  cost  of  plastering  its  walls  and  ceiling,  at  $0.40  per 
square  meter ;  (zz.)  the  cost  of  papering  its  walls  with  paper  8ocm  wide  and 
iom  in  a  roll,  at  $1.75  a  roll  put  on,  and  with  a  gilt  border  next  the  ceiling 
that  costs  $0.20  per  meter. 

66.  A  man  has  a  rectangular  garden  68m  long  and  42"*  wide.     He  make:; 
around  it  a  path  3.4™  wide.     What  area  is  left  for  other  purposes  ? 

67.  A  field  is  6oom  by  240™.     How  much  of  its  area  will  be  taken  by  a 
road  2Om  wide  passing  through  the  middle  and  parallel  to  the  longer  side  ? 

68.  Through  the  middle  of  a  rectangular  garden  i8om  by  76™,  two  paths 
at  right  angles  to  each  other  and  parallel  to  the  sides  are  made,  each  i.4m 
wide.     Find  the  area  (z.)  of  the  paths  ;  (z'z.)  of  what  remains. 

69.  A  mirror  is  2.8m  by  i.9m,  and  the  wooden  frame  is  4cm  wide.    Find  the 
area  of  the  glass  and  also  of  the  frame. 

70.  A  table  2.3m  by  i.2m  is  covered  with  cloth,  which  overlaps  3cm  on  all 
sides.     How  much  cloth  is  needed  ? 

71.  The  bridge  over  the  Elbe  at  Dresden  is  400™  long  and  io.3m  wide. 
How  many  stones  i8cm  by  15°™  are  required  to  cover  the  roadway,  the  foot- 
paths having  together  a  width  of  2.45™? 

72.  How  many  shingles  8in  wide,  laid  6in  to  the  weather,  will  cover  the 
roof  of  a  house,  the  ridge  being  52**  long,  the  slant  side  of  the  roof  24ft,  and 
the  bottom  course  on  each  side  being  double  ? 

§  138.    THE  (oblique)  PARALLELOGRAM  (S  =  a  X  h) . 

In  a  parallelogram,  given,  — 

73.  Base  =  27cm,  altitude  =  22cm ;   find  the  area. 

74.  Base  —  91. 25m,  altitude  =  40™;   find  the  area. 

75.  Base  =  640dkm,  altitude  =  i8odkm;   find  the  area  in  hectars. 

76.  Area  =  84*,  base  =  105™;   find  the  altitude. 

77.  Area  =  iha,  altitude  =  I25m  ;  find  the  base. 

78.  How  far  apart  are  the  sides  of  a  rhomboid  containing  2lO^m,  if  one  of 
these  sides  =  I5m? 

79.  Perimeter  of  a  rhombus  —  6m,  distance  apart  of  two  parallel  sides  = 
75cm;   find  the  area. 

80.  Find  the  area  of  a  rhombus  whose  diagonals  are  1 6m  and  1 2m. 

81.  The  area  of  a  rhombus  —  24O(im,  one  diagonal  =  iom;   find  the  other. 

82.  Find  the  area  of  a  rhombus  if  the  sum  of  the  diagonals  =  2Om  and  their 
ratio  is  3  :  5. 


150  GEOMETRY   FOR   BEGINNERS.  [§   139. 

83.  Construct  a  rhomboid,  and  then  find  its  area. 

84.  Construct  a  rhombus,  and  then  find  its  area. 

85.  Construct  a  rhombus  and  a  square,  both  having  the  same  perimeter. 
Which  has  the  greater  area  ?    Prove  your  answer. 

86.  A  rhombus  and  a  rectangle   have   equal  perimeters.     A  side  of  the 
rhombus  =  5m,  and  a  side  of  the  rectangle  =  3-5m.     Find  the  perimeter  and 
the  two  areas,  if  the  altitude  "of  the  rhombus—  3.5™. 

§  139.    THE  TRIANGLE  (S=^ax  h). 

In  a  triangle,  given,  — 

87.  Base  =  50m,  altitude  =  40™;   find  the  area. 

88.  Base  =  80.5™,  altitude  =  6om;   find  the  area. 

89.  Base  =  2.4km,  altitude  =  i.875km;   ^n(^  tne  area  *n  nectars. 

90.  Area  =  45Oha,  base  =  5km ;  find  the  altitude. 

91.  Area  =  4O(im,  altitude  =  12.5™;   find  the  base. 
In  a  right  triangle,  given,  — 

92.  The  legs  4™  and  5.6™;   find  the  area. 

93.  The  legs  34™  and  25™;   find  the  area  in  ars. 

94.  The  area  =  15*,  one  leg  =  50™;   find  the  other  leg. 

95.  Sum  of  the  legs  =  2Om,  their  ratio  =  2:3;   find  the  area. 

96.  Sum  of  the  legs  =  70™,  and  one  is  6  times  the  other ;   find  the  area. 

97.  Construct  a  triangle,  and  then  find  its  area. 

98.  If  the  base  of  a  triangle  =  50™,  what  must  be  its  altitude  in  order  that 
its  area  may  be  the  same  as  that  of  a  rectangle  whose  sides  are  40™  and  25™? 

99.  Compare  (that  is,  find  the  ratio  of)  the  areas  of  a  square,  and  of  a  tri- 
angle whose  base  is  one  side  of  the  square  and  vertex  is  in  the  opposite  side. 
Does  it  matter  where  in  the  opposite  side  the  vertex  is  ?     Why  not  ? 

100.  Find  the  side  of  a  square  equivalent  to  a  triangle  with  the  base  90™ 
and  the  altitude  20™. 

101.  The  base  of  a  triangle  =  6m,  altitude  =  4™.     Find  the  base  of  a  tri- 
angle with  the  same  altitude  and  twice  as  large  ;   3  times  as  large ;  4  times 
as  large  ;    10  times  as  large. 

102.  Two  triangles  have  the  same  base,  I2m,  and  their  altitudes  are  8m  and 
24™.     Find  the  ratio  of  their  areas. 

103.  Two  triangles  are  each  I5m  high,  and  their  bases  are  20™  and  30™. 
Find  the  base  of  a  triangle  equivalent  to  their  sum. 

104.  From  a  triangle  with  base  =  15™,  and  altitude  =  8m,  a  triangle  with 
the  base  =  6m  is  cut  off  by  a  line  drawn  from  the  vertex.     What  part  of  the 
whole  triangle  is  the  triangle  cut  off  ? 


§    14°-]  CHAPTER   VII. AREAS.  151 

105.  Same  exercise,  making  the  altitude  I2m  instead  of  8m. 

106.  Find  the  entire  surface  of  a  pyramid,  with  a  square  base  (see  page  4, 
Fig.  2,  IV.),  if  a  side  of  the  base  =  2m,  and  the  altitude  of  one  of  the  lateral 
sides  =  4m. 

107.  The  roof  of  a  tower  consists  of  4  isosceles  triangles.     The  base  of 
each  triangle  =  2. 2m,  the  altitude  =  5-4m.     How  many  square  meters  of  tin 
are  required  to  cover  the  roof  ? 

§  140.  THE  TRAPEZOID  (S1  =  £[#  +  c]  X  /*),  AND  THE  TRAPEZIUM. 

108.  Find  the  area  of  a  trapezoid,  if  the  parallel  sides  are  2Ocm  and  3Ocm, 
the  altitude  I2cm. 

109.  Area  of  a  trapezoid  =  3691™,  the  parallel  sides  are  i6m  and  2^m.   Find 
the  altitude. 

110.  Area  of  a  trapezoid  =  I24.8im,  altitude  =  6.4m,  one  of  the  parallel 
sides  =  I2.8m ;  find  the  other. 

111.  A  garden  has  the  shape  of  a  trapezoid.     The  parallel  sides  are  6om 
and  32™,  and  their  distance  apart  I24m.    Find  the  side  of  an  equivalent  square. 

112.  Construct  a  trapezoid,  and  then  find  its  area. 

113.  A  board  is  8.5™  long,  o.45m  wide  at  one  end,  and  o.36m  wide  at  the 
other.     Find  its  value  at  $0.50  per  square  meter. 

114.  A  wall  I2m  long  is  3.15™  high  at  one.end  and  2."jm  at  the  other.    Find 
its  area. 

115.  The  area  of  a  field  shaped  like  a  trapezoid  =  I5ha.    The  parallel  sides 
are  i68.7m  and  144™.     How  far  apart  are  these  sides  ? 

116.  A  court-yard  has  the  shape  of  a  trapezoid.     The  parallel  sides  are 
20.4™  and  18.2™,  and  their  distance  apart  is   i6m.     How  many  stones,  each 
containing  5<idm,  will  be  required  to  pave  the  yard  ? 

117.  What  will  it  cost  to  shingle  a  trapezoidal  roof,  the  parallel  sides 
being  15.8™  and  n.6m,  their  distance  apart   6.2m,  and  the  price  for  shingling 
being  $1.20  per  square  meter  ? 

118.    A  diagonal  in  a  trapezium  =  32™,  and  the  altitudes  of  the  triangles 
made  by  the  diagonal  are  i8m  and  2Om.     Find  the  area  of  the  trapezium. 

119.  Construct  a  trapezium,  and  then  find  its  area. 

120.  The  area  of  a  four-sided  field  =  24ha.     The  altitudes  of  the  triangles 
made  by  a  diagonal  are  140'"  and  158™.    Find  the  length  of  the  diagonal. 

121.  In  a  quadrilateral,  a  diagonal  =  40™,  the  altitude  of  one  triangle  =  I2m, 
and  the  altitude  of  the  other  triangle  is  four  times  as  great.     Find  the  area  of 
the  quadrilateral. 


152  GEOMETRY    FOR   BEGINNERS.  [§    141. 

§  141.    POLYGONS  (if  regular,  S=£#xr). 

The  less  radius  r  of  a  regular  polygon  has  a  fixed  ratio  (found  by  Trigo- 
nometry) to  the  length  of  one  side. 

In  a  regular  triangle,  r  =  one  side  X  0.2887. 
In  a  regular  quadrilateral,  r  —  one  side  X  0.50x30. 

In  a  regular  pentagon,  r=  one  side  X  0.6882. 

In  a  regular  hexagon,  r  =  one  side  X  0.8660. 

In  a  regular  octagon,  r  =  one  side  X  1.2071. 

In  a  regular  decagon,  r  =  one  side  X  1.5388. 

In  a  regular  dodecagon,  r  =  one  side  X  i  .8660. 

122.  One  side  of  a  regular  triangle  —  2ocm;   find  the  area. 

123.  One  side  of  a  square  =  40°™;    find  the  area. 

124.  One  side  of  a  regular  pentagon  =  5Ocm;   find  the  area. 

125.  The  perimeter  of  a  regular  hexagon  =  3™;   find  the  area. 

126.  The  perimeter  of  a  regular  octagon  =  i6m;   find  the  area. 

127.  The  perimeter  of  a  regular  decagon  =  2OOm;   find  the  area. 

128.  One  side  of  a  regular  dodecagon  =  6om;   find  the  area. 

129.  The  area  of  a  regular  hexagon  =&*,  the  less  radius  =  240™;  find  one 
side  and  the  perimeter. 

130.  The  area  of  a  regular  octagon—  i6o<im,  one  side=  5m;   find  the  less 
radius. 

131.  Construct  a  regular  pentagon,  and  then  find  its  area. 

132.  Construct  a  regular  hexagon,  and  then  find  its  area. 

133.  Find  the  side  of  a  square  equivalent  to  a  regular  hexagon  one  side 
of  which  =  4m. 

134.  Find  the  entire  surface  of  a  hexagonal  prism  (see  page  4,  Fig.  2,  III.), 
if  the  altitude  =  i.2m,  and  one  side  of  the  base  =  4Ocm. 

135.  Required  the  total  area  of  the  hexagons  in  Fig.  //j,  if  a  side  of  each 
hexagon  =  36"". 

136.  Compute  the  total  area  covered  by  the  figures  in  Fig.  116,  if  a  side  of 
each  figure  =  1 .4m. 

137.  Compute  the  total  area  covered  by  the  figures  in  Fig.  127,  if  a  side 
of  each  figure  =  2m. 

138.  The  floor  of  a  church  is  30™  by  I5m,  and  is  paved  with  regular  hex- 
agonal stones,  alternately  white  and  black,  a  side  of  each  stone  being  I2cm. 
How  many  stones  of  each  kind  were  required,  two  per  cent  being  added  to 
fill  the  corners  ? 


CHAPTER   VII.  —  AREAS. 


153 


139.  A  garden  in  the  form  of  an  irregular  hexagon  can  be  divided  into  the 
following  triangles :  — 

First  triangle :  base  =  36.6m,  altitude  =  6.6m. 
Second  triangle  :  base  —  42.4™,  altitude  =  2Om. 
Third  triangle  :  base  =  42.4™,  altitude  =  22m. 
Fourth  triangle :  base  =  28.4m,  altitude  —  9.8m. 
How  many  ars  does  the  garden  contain? 

140.  Construct  two  equal  heptagons  ;  then  find  the  area  of  the  first  hepta- 
gon by  one  of  the  methods  illustrated  in  Fig.  128,  and  the  area  of  the  second 
by  the  method  illustrated  in  Fig.  129.     IT  the  results  do  not  agree,  what  is  the 
reason  ? 

IV.  —  Theorem  of  Pythagoras. 

§  142.  If  we  construct  a  right  triangle  (Pig.  131),  having  for  its 
legs  three  and  four  units  of  length,  we  shall  find  that  the  hypote- 
nuse will  contain  exactly  five  of  these 
units  of  length.  If  we  then  construct 
squares  upon  the  three  sides  of  this 
triangle,  and  divide  these  squares  into 
square  units,  as  shown  in  the  figure, 
it  will  be  evident  that  the  square  con- 
structed upon  the  hypotenuse  is  just 
equal  to  the  sum  of  the  squares  con- 
structed upon  the  two  legs ;  for,  the 
former  square  contains  25  square  units, 
and  the  latter  squares  16  and  9  of 
these  units,  respectively. 

This  is  a  simple  example  of  a  very  useful  theorem  called,  from 
the  name  of  the  discoverer,  the  Theorem  of  Pythagoras.1 

Theorem.  —  In  every  right  triangle  the  square  of  the  hypote- 
nuse is  equal  to  the  sum  of  the  squares  of  the  two  legs. 

Several  different  proofs  of  this  theorem  have  been  devised,  two 
of  which  we  shall  give. 


Fig.  131. 


1  Pythagoras,  born  at  Samos  about  570  B.C.,  was  one  of  the  most  renowned  phi- 
losophers and  teachers  of  antiquity. 


154 


GEOMETRY    FOR    BEGINNERS. 


[§ 


Proof  I.  —  Upon  the  hypotenuse  A  C  (Fig.  132}  of  a  right 
triangle  ABC  construct  the  square  ACDE;  from  D  and  E 

let  fall  upon  AB  prolonged  the  per- 
pendiculars DF  and  EG  ;  also  let  fall 
upon  DFi\\Q  perpendiculars  C//and 
EK.  From  this  construction,  it  fol- 
lows that  the  right  triangles  ABC, 
AE^G,  CDff,  and  DEK,  which  taken 
in  order  we  will  call  I.,  II.,  III.,  IV., 
are  equal  (why?)  ;  also,  that  BCHF 
is  equal  to  the  square  upon  BC,  and 
EGFK  to  the  square  upon  AB  (why  ?). 
Now  the  square  ACDE  —  the  figure 
ACHKE  4-  the  triangles  III.  and  IV.;  take  from  the  square 
these  triangles,  and  place  them  in  the  positions  of  the  triangles 

I.  and  II.,  then  it  is  clear  that  the 
square  will  be  transformed  into  the 
figure  GB  CHKE,  which  is  composed 
of  the  squares  BCHF  and  EGFK 
of  the  two  legs.  Hence  the  square  of 
the  hypotenuse  ACDE  must  have 
the  same  area  as  the  sum  of  the 
squares  of  the  legs. 

NOTE.  —  This  proof  can  be  readily  shown 
in  a  material  form  by  drawing  Fig.  132  on 
cardboard,  cutting  out  the  triangles  I.,  II.,  III., 
and  IV.,  and  then  changing  the  places  of  two 
of  them  as  just  explained. 

Proof  //.    (Euclid's^   Proof  ).- 
Upon  the  sides  of  the  right  triangle  ABC  (Fig.  ijj)  construct 

1  Euclid  was  a  Greek  geometer  who  lived  about  300  B.C.  He  collected  the  scat- 
tered geometrical  knowledge  then  existing  into  a  famous  treatise  known  as  Euclid's 
Elements, —  a  book  still  in  use,  but  as  unsuited  for  beginners  as  it  is  admirable  for 
rigor  of  logic,  and  for  showing  on  how  few  axioms  the  science  of  Geometry  can  be 
made  to  depend. 


// 


§    1 43.]  CHAPTER   VII. AREAS.  155 

squares,  and  draw  BMA.AC.  Join  BD  and  C F.  &ABD  ^ 
A CF  (II.  Law  of  Equality) .  &ACF has  the  same  base  AF  and 
the  same  altitude  AB  as  the  square  ABFG;  therefore  (§  130) 
A  A CF=  %A£2.  And  &ABD  has  the  same  base  AD  and  the 
same  altitude  DM  as  the  rectangle  ADMN;  therefore  &ABD 
=  £  rectangle  ADMN.  But  if  the  halves,  here  the  triangles 
A  CF  and  AB D,  are  equal,  the  wholes  must  also  be  equal  (Axiom 
II.)  ;  therefore  AB*  =  rectangle  ADMN.  In  the  same  way  (by 
joining  A K  and  BE),  it  can  be  proved  that  BC*  =  rectangle 
CE MN.  Therefore  AB*  +  B C~  =  the  sum  of  the  rectangles 
ADMN  and  CEMN =  the  square  of  the  hypotenuse  AC. 

Corollaries.  —  I.  If  from  the  square  of  the  hypotenuse  we 
subtract  the  square  of  either  leg,  the  remainder  will  be  equal  to  the 
square  of  the  other  leg. 

2.  If  a  and  b  denote  the  legs,  c  the  hypotenuse,  of  a  right  triangle, 

<2=a2  +  t2,     a^c2-?,     and    P^^-a2. 

3.  A  useful  corollary  follows  from  the  second  proof,  namely : 
If  we  construct  squares  upon  the  sides  of  a  right  triangle,  and  from 
the  vertex  of  the  right  angle  draw  a  line  perpendicular  to  the 
hypotenuse,  this  line  will  divide  the  square  of  the  hypotenuse  into 
two  rectangles,  each  equal  to  the  square  of  the  adjacent  leg. 

§  143.    PRACTICAL  EXERCISES  AND  APPLICATIONS. 

In  a  right  triangle,  given,  — 

1.  The  legs  =  24™  and  =  7™;   find  the  hypotenuse. 

2.  The  legs  =  35™  and  —  12™;   find  the  hypotenuse. 

3.  The  legs  =  45™  and  —  28™;   find  the  hypotenuse. 

4:.  The  legs  =  4.7™  and  =  1.104™;   find  the  hypotenuse. 

5.  The  legs  =  72™  and  =  i8m;    find  the  hypotenuse. 

6.  The  legs  =  2dm  and  —  15°™;   find  the  hypotenuse. 

7.  The  hypotenuse  =  65™,  one  leg=  33m ;   find  the  other  leg. 

8.  The  hypotenuse  =  89™,  one  leg—  8om;   find  the  other  leg. 

9.  The  hypotenuse  =  65™,  one  leg  =  63m;   find  the  other  leg. 
10.  The  hypotenuse  =  42.6"%  one  leg  =  I94m;   find  the  other  leg. 


156  GEOMETRY   FOR   BEGINNERS.  [§    143. 

11.  How  long  must  a  ladder  be  to  reach  the  top  of  a  wall  6m  high,  if  the 
foot  of  the  ladder  is  3.2™  from  the  bottom  of  the  wall  ? 

12.  A  ladder  13™  long  leans  against  the  top  of  a  wall,  its  foot  being  5™ 
from  the  bottom  of  the  wall.     How  high  is  the  wall  ? 

13.  The  slant  height  of  a  roof  =  14™,  its  vertical  height  =  6.2m.     Find  the 
width  of  the  building. 

14.  A  triangle  with  the  sides  3,  4,  5,  is  a  right  triangle.     What  kind  of  a 
triangle  is  one  with  the  sides  6,  8,  10  ?  with  the  sides  9,  12,  15  ?     Give  other 
instances. 

15.  If  the  perimeter  of  a  right  triangle  =  96m,  and  the  sides  are  to  each 
as  3  :  4  :  5>  find  the  sides  (see  page  50,  Exercise  1 8). 

16.  If  the  legs  of  a  right  triangle  are  as  3:4,  and  the  hypotenuse  =  40™, 
find  the  legs 

17.  A  rope  41™  long  is  tied  by  one  end  to  the  top  of  a  mast,  and  by  the 
other  end  to  a  ring  on  the  deck  9m  from  the  foot  of  the  mast.     How  high  is 
the  mast  ? 

18.  The  longest  side  of  a  meadow  in  the  shape  of  a  right  triangle  cannot 
be  measured  directly,  by  reason  of  a  swamp.     Compute  its  length,  if  the  other 
sides  are  8o.6m  and  i6om. 

19.  How  many  meters  of  wall  must  be  built  to  fenc'e  in  a  garden  in  the 
shape  of  a  right  triangle,  the  legs  being  1 2Om  and  35™  ? 

20.  A  garden  has  the  shape  of  a  right  triangle.     The  hypotenuse—  i3Om, 
one  leg  =  87.5™.     Find  (z.)  the  area  of  the  garden;    (w.)  the  distance  from 
the  right  angle  to  the  hypotenuse. 

21.  A  cannon-ball,  fired  into  the  air  at  an  angle  of  45°  with  the  horizon, 
passes  over  345™  in  one  second.     How  high  is  it  at  the  end  of  the  second  ? 

22.  Two  travellers  start  together  from  the  same  place  and  travel,  each  4 
miles  an  hour,  one  due  north,  the  other  due  east.    When  will  they  be  60  miles 
apart  ?     How  far  apart  will  they  be  after  each  has  travelled  80  miles  ? 

23.  Find  the  area  of  an  isosceles  triangle,  if  one  of  the  equal  sides  =  iom 
and  the  base  =  8om  (see  §  80). 

24.  In  an  isosceles  triangle,  the  base  =  i8.4m,  the  perimeter  —  64™.     Find 
the  area. 

25.  Prove  that  in  an  isosceles  right  triangle  the  altitude  is  equal  to  half 
of  the  base. 

26.  The  area  of  an  isosceles  right  triangle  =  84<im;   find  the  base. 

27.  The  altitude  of  an  isosceles  right  triangle  =  2m;  find  the  side  of  an 
equivalent  square. 

28.  In  an  ordinary  roof  the  two  sides  are  I2m  apart  at  the  bottom,  and  the 
vertical  height  is  4.5™.     Find  the  length  of  the  rafters. 


§  143']  CHAPTER  VII. AREAS.  1&7 

29.  The  side  of  an  equilateral  triangle  =  20™ ;   find  the  altitude  and  the 
area  (see  §  80) . 

30.  If  the  altitude  of  an  equilateral  triangle  =  15™,  find  one  side. 
Hint. — The  altitude,  one  side,  and  half  the  base,  form  a  right  triangle. 

31.  If  a  —  one  side  of  an  eqilateral  triangle,  prove  that  the  altitude  =  \  ^'s/3 
and  the  area  =  \  #2A/3- 

32.  A  garden  in  the  shape  of  an  equilateral  triangle  contains  6cx3^m;   find 
the  side, 

33.  Find  the  side  of  a  square  equivalent  to  an  equilateral  triangle  whose 
side  =  86m. 

34.  A  square  contains  I2ha.     Find  the  side  of  the  equivalent  equilateral 
triangle. 

35.  The  side  of  a  square  —  6m.     Find  its  diagonal. 

36.  What  is  the  side  of  a  square  if  the  diagonal  =  8m? 

37.  The  diagonal  of  a  rectangle  =  o.86m,  one  side  =  0.3™.     Find  the  other 
side. 

38.  The  diagonals  of  a  rhombus  are  4™  and  6m.     Find  its  perimeter. 

39.  How  far  apart  are  the  opposite  corners  of  a  floor  I2m  by  i8m? 

40.  What  must  be  the  least  thickness  of  a  log  from  which  a  rectangular 
beam  24cm  by  40°™  can  be  obtained  ? 

41.  How  much  is  the  diagonal  of  a  square  increased  if  the  side  is  doubled? 
trebled  ?  quadrupled  ? 

42.  How  much  is  the  side  of  a  square  increased  if  the  diagonal  is  doubled? 
trebled?  increased  tenfold? 

43.  Find  the  less  radius  of  a  regular  hexagon  if  the  perimeter—  icom. 

44.  Compute  the  side  of  a  square  equivalent  to  the  sum  of  the  squares  whose 
sides  are  12™  and  i6m. 

45.  Construct  a  square  equivalent  to  the  sum  of  two  given  squares. 

46.  The  side  of  a  square  =  2Ocm.    Construct  a  square  twice  as  large.    Find 
the  length  of  its  side,  (z.)  by  measurement ;    (zV.)   by  computation. 

47.  Construct  a  square  equivalent  to  the  sum  of  the  squares  whose  sides 
are  3Ocm,  4Ocm,  5ocm. 

48.  The  side  of  a  square—  i6cm.     Construct  a  square  three  times  as  large. 

49.  Construct  a  square  equivalent  to  the  difference  of  two  given  squares 
(apply  §  74). 

50.  Construct  a  rectangle  such  that  the  square  upon  its  diagonal  shall  be 
five  times  the  square  upon  one  side. 


158  GEOMETRY   FOR  BEGINNERS.  [§   144. 

V.  —  Transformation  of  Figures. 

§  144.  Definition.  —  To  TRANSFORM  a  figure  is  to  change  its 
shape  without  changing  its  size  ;  in  other  words,  to  change  it  to  an 
equivalent  figure. 

What  instance  of  this  kind  of  change  was  given  in  §  129  ? 

There  are  two  ways  of  effecting  these  changes  :  — 

1.  By  computation.     Compute  such  parts  of  the  new  figure  as 
are  necessary  to  determine  it ;  we  can  then  construct  it,  if  we  wish 
to  do  so.     Examples  of  this  method  occur  among  the  Exercises 
in  Parts  III.  and  IV.  of  the  present  chapter  (see  Part  III.,  Exer- 
cises 1 8,  46,  47,  100,  133,  and  Part  IV.,  Exercises  27,  33,  34). 

2.  By  direct  construction.     This  was  the  method  employed  in 
§  129,  and  a  few  more  useful  examples  will  now  be  added. 

§  145.   Problem.  —  To  transform  a  triangle  ABC  (Fig.  134) 
into  an  isosceles  triangle. 

Analysis.  —  The   vertices   of   all    triangles 
equivalent  to  ABC,  and  having  AB  as   a 
base,  must  lie  in  the  line  CD  \\AB($  130). 
Of  these  triangles,  that  which  has  its  vertex  E 
equidistant  from  A  and  B  will  be  isosceles. 

B          Hence  give  the  construction  required. 
Fig- 134- 

Exercises.  —  1.    Construct  a  triangle,   and   then 

transform  it  into  sen  isosceles  triangle. 

2.  Transform  a  given  triangle  into  a  right  triangle. 

3.  Transform  a  given  triangle  into  another  having  the  angle  60°. 
4»   Transform  a  given  parallelogram  into  a  rectangle. 

5.   Transform  a  given  parallelogram  into  another  having  the  angle  45°. 

§  146.  Problem.  —  To  transform  a  triangle  ABC  (Fig.  135) 
into  another  having  a  given  base  b. 

Analysis.  —  If  on  AB  prolonged  we  take  AD  =  b,  and  join 
CD,  A  A  CD  will  have  the  given  base  b,  but  will  be  too  large  by 


§  I47-] 


CHAPTER  VII.  —  AREAS. 


159 


*35> 


the  &BCD.    Therefore  construct  the  &CED  ^  &BCD  (§  130, 

Corollary) ,  and  take  it  from  A  A  CD,      f_ 

leaving  A  AE  D,  which  is  the  trian- 
gle required. 

Give  in  full  the  construction  re- 
quired. 

Exercises.  —  1.  Solve  this  problem  for 
the  case  where  the  given  base  b  is  less  than 
the  base  of  the  given  triangle. 

2.  Construct  a  triangle  with  the  sides  4dm,  6*«,  and  8dm,  and  then  trans- 
form it,  — 

(z.)     Into  an  isosceles  triangle  with  the  base  5dm. 
(z'z.)    Into  a  right  triangle  with  a  leg  equal  to  4dm. 
(z'z'z.)  Into  an  equilateral  triangle, 
(zz/.)  Into  a  triangle  with  the  angle  50°  and  the  base  3dm. 

3.  Transform  a  given  parallelogram  into  another  having  a  given  side. 

§  147.   Problem.  —  To  transform  a  triangle  ABC  (Fig.  ij6) 
into  another  having  a  given  altitude  h. 
Analysis   and   construction   to   be 
given  by  the  learner   (with  the  aid 
of  Fig.  ij6). 

Exercises.  —  1.  Solve  this  problem  for 
a  case  in  which  the  new  altitude  h  is  less 
than  that  of  the  given  triangle. 

2.   Construct  a  triangle  with  the  sides 


Fig.  136. 


4dm,  6dm,  and  8dm,  and  transform  it,  — 

(z.)    Into   a  triangle,  the  altitude 


(z'z.)  Into  a  triangle  with  the  angle 

70°  and  the  altitude  4dm.       E 

§  148.   Problem.  —  To  transform 
a  triangle  ABC  (Fig.  137)   into  a    A 
rectangle. 


Fig.  137. 


Construction.  —  At  A  and  B  erect  perpendiculars  to  AB,  and 


160 


GEOMETRY   FOR   BEGINNERS. 


[§   149' 


through   Z>,  the    middle   of  A  C,  draw  EF  II  AB.      Rectangle 
ABEF  =  &ABC. 

Proof.  —  Show  that  each  is  half  of  the  rectangle  ABHG. 
.     Exercises.  —  1.   Transform  a  triangle  into  a  parallelogram  with  the  an- 
gle 60°. 

2.  Construct  a  rectangle  with  the  sides  6dm  and  8dm,  and  then  transform 
it  into  (?'.)  a  triangle  ;  («'.)  a  right  triangle ;  (m.)  an  isosceles  triangle  ; 
(z'z/.)  an  equilateral  triangle. 


§  149.    Problem.  —  To  transform  a  rectangle  ABCD  (Fig. 
138)  into  a  square. 

Construction.  —  Make  BE  =  BC, 
and  about  O,  the  middle  point  of  B  E, 
describe  an  arc  cutting  DA  prolonged 
in  F.  Join  FB,  and  the  square  con- 
structed upon  FB  will  be  equal  in  area 
to  the  rectangle  ABCD. 

Proof.  —  EBF  is  a  right  triangle. 
For  the  triangles  OEF  and  OBF  are 
isosceles  (why?)  ;  hence EFO  =  FEO, 
and  OFB  =  OBF  (§  80).  By  addi- 
tion, EFO  +  OFB  =  EFB  =  FEO 
-f  OBF.  But  EFB  +  FEO  +  OBF 
=  180°  (§  64)  ;  therefore  EFB  =  90°. 


l&L. 


Fig.  138. 


Since  EBFis  a  right  triangle,  it  follows  from  §  142,  Corollary  3, 
that  ~FJ?  =  the  area  of  the  rectangle  ABCD. 

Exercises.  —  1.  Construct  a  rectangle  with  the  sides  9dm  and  4dm,  and 
transform  it  into  a  square.  Measure  the  side  of  the  square.  What  is  its 
length  ?  What  ought  its  length  to  be  ? 

2.  Construct  a  triangle  with  the  sides  4dm,  6dm,  and  8dm,  and  transform  it 
into  a  square. 

3.  Transform  a  given  parallelogram  into  a  square. 

§  150.  Problem.  —  To  transform  any  polygon  ABCDE  (Fig. 
139}  int°  a  triangle. 


§     I51-]  CHAPTER   VII.—  AREAS.  161 

Construction.  —  Draw  a  diagonal  AD.  Through  E  draw  a  line 
parallel  to  AD  and  meeting 
A  B  prolonged  in  F.  Join 
DF.  Then  is  the  quadri- 
lateral BCDF  equivalent 
to  the  pentagon  ABODE. 

Repeat  this  construction 
by  drawing  the  diagonal 
JSD,  a  parallel  through  C, 
and  then  joining  DG ;  this 

reduces    the     quadrilateral  F&*39* 

BCDE  to  the  equivalent  triangle  FGD. 

Proof.  —  Left  for  the  learner,  with  the  aid  of  Fig.  ijp  and  the 
analysis  of  §  146. 

Corollary.  —  Hence,  any  polygon  may  be  transformed  into  a 
square.  By  what  three  steps  ? 

This  supplies  another  way  of  finding  the  area  of  a  polygonal  field. 
Construct  (to  scale)  the  polygon  on  paper,  transform  it  into  a 
square,  and  multiply  one  side  of  this  square  by  itself. 

Exercises.  —  1.   Transform  a  hexagon  into  a  triangle. 

2.  Transform  a  pentagon  into  a  square. 

3*  Construct  three  equal  octagons.  Find  the  area  of  the  first  and  second 
by  two  of  the  methods  given  in  §  134,  and  the  area  of  the  third  by  transform- 
ation into  a  square. 

VI.— Partition  of  Figures. 

§  151.  In  buying  and  selling  land  it  often  becomes  necessary 
to  run  lines  of  division  through  estates,  forests,  fields,  etc.,  dividing 
them  into  smaller  parts  in  some  assigned  way. 

This  is  done  either  (/.)  by  computation  (arithmetical  partition) 
or  («".)  by  construction  (geometrical  partition) . 

The  first  method,  when  practicable,  is  generally  adopted.  Let 
us  examine  some  cases. 


162 


GEOMETRY   FOR    BEGINNERS. 


[§   152. 


§  152.    TRIANGLES.     Case  1.  —  A  line  CD  (Fig.  140)  joining 
the  vertex  of  a  triangle  ABC  to  the  middle  point  of  the  base  AB 
divides  the  triangle  into  two  equal  parts 
(§  130,  Corollary). 

Again,  it  AE=.  \AB,  and  EB  =  \ 
AB,  then  &ACE  =  \kABC,  and 
&ECB  =  f  &ABC  (why?).  And, 
A  ACE:&  ECB  =  1:2.  In  gen- 
eral, — 

A  line  joining  the  vertex  of  a  triangle  to  the  base  divides  both 
the  triangle  and  its  base  into  parts  which  have  the  same  ratio. 

What  ratio  have  the  triangles  A  EC  and  ABCt  the  triangles 
ECBvrAABCt 

Case  2.  —  Line  of  division  DE  (Fig.  141)  passes  through  two 
sides  of  the  triangle. 

=  I  AC]  then 
i),and &ADE 

=  $  A  ADC  (Case  i)  ;   therefore  A  ADE 
=  J  X  \&ABC=  %t\ABC. 

Again,  if  CE  =  §  CA,  and  CF=%  CB, 
then  (by  the  same  reasoning)  A  CEF  —  \ 
X  %&ABC^  %&ABC;  that  is,  A  CEF 
\&ABC=2\*i.     In  general, — 
The  ratio  of  the  triangle  cut  off  to  the  entire  triangle  is  equal  to 
the  product  of  the  ratios  between  the  parts  of  the  sides  that  meet 
at  the  common  vertex. 

Case  3.  —  Hence,  if  a  definite  part  —  say,  one-fourth  —  of  a  tri- 
angle is  to  be  cut  off  by  a  line  from  side  to  side,  this  can  be  done 
in  various  ways,  because  the  fraction  £  may  be  obtained  as  the 
product  of  various  pairs  of  factors;  for  example,  £  and  J,  \  and  |, 
f  and  |,  I-  and  ft,  etc. 

If,  however,  the  line  of  division  has  to  pass  through  a  giver) 
point  in  one  side,  then  its  position  is  determined. 


B 


F 
Fig.  141. 


§    1 5 3-]  CHAPTER   VII. AREAS.  163 

For  example,  if  AD  (Fig.  142}  =  \AB,  and  the  AADE  is 
to  be  made  equal  to  \AABC,  then  the  line 
of  division  will  pass  through  a  point  in  the 
side  AC,  since  A  ADC  would  be  too  large 
(how  large  is  it?) ;  and,  since  |  -j-  §  =  |,  the 
line  will  pass  through  a  point  E  such  that 


How  must  a  line  DF  be  drawn  if  A  B DF 
is  to  be  made  equal  to  \AABC1 

§  153.   PARALLELOGRAMS.     Case  1.  —  Line  of  division  joins  two 
opposite  corners.     See  §  102. 

Case  2.     Line  of  division  AE  (Fig.  143)  joins  a  corner  to  a 
point  in  one  side. 

Let  (Fig.  143}  CE  =  J.  CD,  then 
A  ACE  =  \AACD  (§  152,  Case  i). 
But  AACD  =  ±  the  parallelogram 

ABCD    (§  102);"  therefore  A  ^  C£ 

Figm I43^        ~~&     =  %  the  parallelogram  ABCD. 

CE   must   be   what  part   of   C.Z2,  if 
A  ACE  =  f  the  parallelogram? 

Case  3.    Line  of  division  joins  two  opposite  sides  and  is  parallel 
to  the  other  sides.     This  case  is  left  for  the  learner  to  investigate. 
How  can  a  rectangle  be  divided  into  six  equal  parts  ? 
Case  4.     Line  of  division  EF  (Fig.  144)  joins  two  opposite 
sides,  and  is  not  parallel  to  the  other  sides. 
Let  (Fig.  144)  CE  =  ±  CD,  AF  — 
%AB,  then  AACE  =  \AACD,  and 
A  AEF  =  AADF  (why?)  =  §  AABD ; 
therefore  A  A  CE  -f-  A  A  E  F  =  the  trap- 

*^P«  ^•44'* 
—  \  of  half  the  parallelogram  -f  f  of 

half  the  parallelogram  =  J  the  parallelogram. 


164 


GEOMETRY    FOR    BEGINNERS. 


[§    154. 


Case  5.  Line  of  division  FE  (Fig.  145)  joins  two  adjacent 
sides.  Let  AE  =  £  AB,  AF  =  f  AC, 
then  (§  152,  Case  2)  &AEF  =  £  X  f 
X  &ABC=  ±AA£C=  TV  the  whole 
parallelogram. 


A  B 

Fig.  i 45. 


§  154.  We  will  now  give  two  or  three 
cases  in  which  the  partition  is  effected 
by  direct  construction.  The  proofs  of  the  constructions  are  left 
to  the  ingenuity  of  the  learner. 

Problem  I.  —  To  divide  a  triangle  ABC  (Fig.  146)  from  a 
point  D  in  one  side  into  two  equal  parts. 

Construction.  —  }^  CD,  bisect  AB  in  E,  draw  EF  II  CD, 
join  DF,     DFis  the  line  of  division  required. 


D       E 

Fig.  146. 


Problem  II.  —  To  divide  a  triangle  ABC  (Fig.  147)  from  a 
point  D  in  one  side  into  three  equal  parts. 

Construction. — Join  CD,  trisect  AB  in  E  and  F,  draw  EG 
and  FH  both  II  CD,  join  DG  and  D H ";  these  last  are  the  lines 
of  division  required. 

Problem  III.  —  To  divide  a  trapezoid  A  B  C  D  (Fig.  148)  from 
<f       H   D          a  point  E  in  one  of  the  parallel  sides  into 
two  equal  parts. 

Construction. — Bisect  the  parallel  sides 
in   F  and  G,   make    GH  =  EF  and 
join  EH:  it  is  the  line  of  division  re- 
Fi/.i48.  quired. 


§   1 55-]  CHAPTER   VII.  —  AREAS.  165 

Will  this  construction  divide  a  parallelogram  into  two  equal 
parts  ?  See  §  153,  Case  4. 

How  is  this  problem  to  be  solved,  if  the  line  of  division  is  to 
start  from  a  given  point  in  one  of  the  parallel  sides  ? 

§  155.     Practical  Exercises. 

1.  Divide  a  triangle  into  five  equal  parts. 

2.  Divide  a  triangle  into  two  parts  having  the  ratio  4 : 7. 

3.  Divide  a  triangle  into  three  parts  which  shall  be  to  each  other  as  the 
numbers  I,  3,  5. 

4r.  Through  one  corner  of  a  triangle  draw  a  line  cutting  off  a  part  equal 
to  f  the  entire  triangle. 

5.  How  can  a  triangle  containing  i2Oim  be  divided  into  parts  containing 
30<im,  40im,  and  -jo<im? 

6.  From  a  triangular  field  worth  $1200  cut  off  a  piece  worth  $400. 

7.  Through  a  triangular  meadow,  a  ditch  runs  from  one  corner  to  a  point 
in  the  opposite  side,  264^  from  one  end,  627yds  from  the  other.     Find  the 
ratio  of  the  two  parts  of  the  meadow  to  each  other. 

8.  A  piece  of  woodland  in  the  shape  of  a  triangle  cost  $6000,  but  is  now 
worth  50  per  cent  more  than  at  first.     It  is  required  to  run  a  line  from  one 
corner  to  the  opposite  side,  which  is   1200  feet  long,  cutting  off  a  portion 
worth  $1800. 

9.  A  field  has  the  shape  of  an  isosceles  right  triangle,  the  equal  sides 
being  each  400™  long.     Through  the  vertex  of  the  right  angle  draw  a  line 
cutting  off  a  part  which  shall  contain  I  hectar. 

Through  two  sides  of  a  triangle  draw  a  line  cutting  off,  — 
10.    |  of  the  triangle.  11.    f  of  the  triangle. 

12.    f  of  the  triangle.  13.    f  of  the  triangle. 

What  ratio  does  the  portion  cut  off  bear  to  the  whole  triangle,  if  the  line  of 
division  cuts, — 

14.  |  from  one  side  and  \  from  the  other  ? 

15.  f  from  one  side  and  f  from  the  other  ? 

16.  f  from  one  side  and  T37  from  the  other  ? 

17.  The  legs  of  a  right  triangle  are  24°™  and  40°™.     Find  the  area  of  the 
part  cut  off  by  a  line  which  passes  through  points  in  the  legs  distant,  respec- 
tively, 6cm  and  iocm  from  the  vertex  of  the  right  angle.     What  is  the  ratio  of 
its  area  to  that  of  the  entire  triangle  ? 


166  GEOMETRY   FOR   BEGINNERS.  [§   155. 

18.  Through  the  middle  point  in  one  side  of  a  triangle  draw  a  line  dividing 
it  into  parts  which  shall  have  the  ratio  3:5. 

19.  Through  the  middle  point  of  one  side  of  a  triangle  draw  lines  dividing 
the  triangle  into  parts  which  shall  be  to  each  other  as  1:2:3. 

20.  A,  B,  and  C  buy  a  triangular  field  ABC  (Fig.  149).     A  pays  $200; 

B,  $500;  C,  $300.  They  wish  to  divide  it  among 
themselves,  by  lines  starting  at  a  point  D,  where 
there  is  a  valuable  spring;  A  to  take  the  upper  part, 
B  the  middle,  and  C  the  lower.  Show  how  the 
lines  of  division  should  be  run  if  A  D  =  1 8om,  D  C 
—  300™. 

21.    Divide  two  sides  of  a  triangle,  each  into  five 
equal  parts,  and  join  in  order  the  points  of  division. 
Fig.  i4Q.  Then  find  the  ratio  of  each  of  the  smaller  triangles 

thus  formed  to  the  entire  triangle. 

22.  What  part  of  a  parallelogram  is  the  triangle  cut  off  by  a  line  drawn 
from  one  corner  to  the  middle  of  a  side  ? 

23.  Cut  off  f  of  a  parallelogram  by  a  line  drawn  through  one  corner. 

24.  From  a  parallelogram  containing  6ooim  cut  a  triangle  containing  2$o<im. 

25.  Divide  a  rectangle  whose  sides  are  36™  and  2Om,  by  lines  drawn  from 
one  corner,  into  three  parts  which  shall  be  to  one  another  as  2:3:4. 

26.  Divide  a  parallelogram  into  seven  equal  parts. 

27.  Divide  a  parallelogram  into  two  parts  having  the  ratio  3 :  5. 

28.  What  part  of  a  parallelogram  lies  between  J  of  one  side  and  f  of  the 
opposite  side  ? 

29.  A  rectangular  field  is  500™  by  300™.    A  tree  is  in  the  longer  side,  150™ 
from  one  corner.     Draw  a  line,  beginning  at  the  tree,  which  shall  cut  from 
the  field  I  hectar,  adjoining  the  150™  part  of  the  longer  side. 

30.  A  line  joining  two  adjacent  sides  of  a  parallelogram  cuts  off  a  triangle 
whose  sides  are  ^  and  f  of  the  corresponding  sides  of  the  parallelogram. 
What  part  of  the  latter  is  the  triangle  ? 

31.  Divide  a  trapezoid  into  five  equal  parts. 

32.  In  a  trapezoid  A  BCD  the  parallel  sides  are  A  B  —  25™,  CD  =  3$m; 
their  distance  apart  =  2Om.     Divide  the  trapezoid  into  two  equal  parts  by  a 
line  beginning  at  a  point  E  in  A  B,  such  that  AE=  iom. 

33.  In  a  quadrilateral  A  BCD,  the  diagonal  BC—  40™,  the  altitudes  of  the 
triangles  ABC  and  BCD  are  2Om  and  30™.     Divide  the  quadrilateral  by  a 
line  through  B  into  two  equal  parts. 


CHAPTER   VII. REVIEW.  167 

REVIEW   OF   CHAPTER  VII. 
SYNOPSIS. 

1.  Surfaces  are  measured  by  choosing  a  unit  and  finding  how  often  this  unit 
is  contained  in  the  surface  to  be  measured. 

2.  The  area  of  a  surface  is  the  number  of  times  the  unit  is  contained  in  the 
surface  followed  by  the  name  of  the  unit. 

3.  Units  of  area  (except  the  acre)  are  squares  whose  sides  are  linear  units. 

4.  The  ratio  between  two  units  of  area  is  the  square  of  the  ratio  between 
the  corresponding  linear  units. 

5.  Geometry  enables  us  to  measure  a  surface  indirectly  by  showing  that  its 
area  depends  upon  the  length  of  certain  lines. 

6.  Area  of  a  square  —  square  of  one  side. 

7.  Area  of  a  rectangle  —  its  base  X  its  altitude. 

8.  A  parallelogram  =  a  rectangle  of  the  same  base  and  altitude  ;  its  area  = 
its  base  X  its  altitude. 

9.  Parallelograms  with  equal  bases  and  equal  altitudes  are  equivalent. 

10.  Area  of  a  rhombus  =  \  product  of  its  diagonals. 

11.  A  triangle  =  \  a  parallelogram  of  same  base  and  altitude;  its  area  =  its 
base  X  half  its  altitude. 

12.  Triangles  with  equal  bases  and  equal  altitudes  are  equivalent. 

13.  Area  of  a  trapezoid  =  \  sum  of  parallel  sides  X  altitude. 

14.  Area  of  a  regular  polygon  =  \  perimeter  X  less  radius. 

15.  Area  of  any  polygon  may  be  found  by  dividing  it  into  triangles  (two 
ways)  or  into  triangles  and  trapezoids. 

16.  In  every  right  triangle  the  square  of  the  hypotenuse  is  equal  to  the  sum 
of  the  squares  of  the  two  legs.     (Two  proofs.) 

17.  The  perpendicular  let  fall  from  the  vertex  of  the  right  angle  upon  the 
hypotenuse  divides  the  square  of  the  hypotenuse  into  two  rectangles, 
equal  respectively  to  the  squares  of  the  adjacent  legs. 

18.  A  figure  may  be  transformed  into  another  figure,  either  (?'.)  by  computing 
such  parts  of  the  new  figure  as  will  enable  us  to  construct  it,  or  («'.)  by 
direct  construction. 

19.  A  square  may  be  found  equivalent  to  any  polygon  by  first  changing  the 
polygon  to  an  equivalent  triangle,  then  the  triangle  to  an  equivalent 
rectangle,  and  then  the  rectangle  to  an  equivalent  square. 

20.  The  division  of  figures  into  parts  is  effected,  either  (*'.)  by  arithmetical 
partitiont  or  (zV.)  by  geometrical  construction. 


168 


GEOMETRY    FOR    BEGINNERS; 


21.  A  line  through  one  corner  of  a  triangle  divides  the  triangle  and  the 
opposite  side  into  parts  having  the  same  ratio. 

22.  If  a  line  is  drawn  between  two  sides  of  a  triangle  cutting  off  fractional 
parts  of  the  sides,  the  product  of  these  fractions  give  the  fractional  part 
which  the  triangle  cut  off  is  of  the  whole  triangle. 

23.  Nos.  21  and  22  enable  us  to  divide  parallelograms  and  polygons  in 
various  ways. 

EXERCISES. 

1.  Compute  the  areas  in  square  meters  (as  given  below,  in  the  last  column 
to  the  right)  of  the  following  regular  polygons,  one  side  of  the  polygon 
in  each  case  being  supposed  to  be  I  meter :  — 


2. 


6. 


Polygon 
of 

One  Side. 

Less  Radius. 

Area. 

3  sides 

0.2887 

0-433° 

4  sides 

0.5000 

I.OOOO 

5  sides 

0.6882 

1.7205 

6  sides 

0.8660 

2.5980 

8  sides 

1.2071 

4.8284 

10  sides 

I-S388 

7.6942 

12  sides 

l.866o 

11.1961 

Same  exercise,  one  side  of  each  polygon  being  taken  as  2  meters. 
Hint.  —  For  the  first  case  (3  sides),  the  less  radius  =2  X  0.2887,  and  the  area 
_  3  X  2  X  2  X  0.2887  _  z  732qm  _  4  times  the  area  (0.433)  found  above  for 

the  same  polygon  when  the  side  is  im. 

3.  Same  exercise,  one  side  of  each  polygon  being  taken  as  3  meters. 
Hint.— In  the  first  case  (3  sides),  the  area  =  3  X  3  X  3  X  0.2887  _  g  times 

2 

the  area  (0.433)  of  the  same  polygon  if  the  side  is  jm. 

4.  From  the  above  exercises  derive  a  general  rule  for  finding  the  area  of  a 
regular  polygon  whose  side  contains  any  number  (a)  of  linear  units. 

5.  Two  fields  of  equal  value  per  unit  of  area  have  the  shape,  one  of  a  square, 
the  other  of  a  regular  hexagon.     If  a  side  of  each  field  =  136™,  and  the 
first  field  is  worth  $600.00,  what  is  the  second  field  worth  ?     What  is  the 
value  per  square  meter  ?  per  hectar  ? 

Government  lands  are  usually  divided  into  tracts  or  townships  6  miles 
square  ;  these  are  subdivided  into  36  sections,  and  each  section  into  half- 
sections,  quarter-sections,  half  quarter-sections,  and  quarter  quarter-sec- 


CHAPTER  VII.  —  REVIEW.  169 

twns.     Find  the  area  in  acres  of  a  township  and  of  each  of  its  sub- 
divisions. 

7.  A  quarter-section  is  fenced  with  a  six-railed  fence,  the  posts  being  I2ft 
apart.     The  rails  cost  $40.00  per  thousand,  the  posts  $60.00  per  thou- 
sand.    Find  the  cost  of  the  fence. 

8.  Of  two  kinds  of  paper,  equal  in  quality,  the  first  is  42°™  by  3Ocm,  and 
costs  $0.20  a  quire  ;   the  second  is  6ocm  by  4Ocm,  and  costs  $0.28  a  quire. 
Which  kind  is  the  cheaper  ? 

9.  A  man  buys  a  rectangular  garden  140™  by  8om.     He  surrounds  the  gar- 
den by  a  wall  o.8m  thick,  and  within  the  wall  by  a  walk  i.8m  wide  ;   and 
through  the  middle  of  the  garden  he  makes  paths  parallel  to  the  sides, 
each  1.5™  wide.     How  much  land  remains  for  other  uses  ? 

10.  A  piece  of  land  is  160™  square.     Another  piece  of  the  same  size  has  the 
shape  of  a  rectangle  four  times  as  long  as  it  is  wide.    Find  the  difference 
between  their  perimeters. 

11.  The  perimeter  of  a  right  triangle  =  6om,  the  hypotenuse  =  25™.   Find  the 
legs  and  the  area. 

12.  The  pejimeter  of  an  isosceles  triangle  =  24™,  the  base  =  6m.     Find  the 
area. 

In  an  isosceles  right  triangle, — 

13.  The  hypotenuse  =  8m;  find  the  area. 

14.  One  leg  =  32™;   find  the  area. 

15.  The  area  =  3Ooim;  find  the  sides. 
In  an  equilateral  triangle, — 

16.  One  side  =  I2to;  find  the  area. 

17.  The  altitude  =  6m;  find  the  area. 

18.  The  area  =  1441™;  find  the  side  and  the  altitude. 

19.  One  side  of  a  square  =  I4m;   find  the  diagonal. 

20.  The  diagonal  of  a  square  =  i6m;   find  one  side. 

21.  In  a  rhombus,  a  diagonal  =  one  side  =  5m ;  find  the  area  of  the  rhombus. 

22.  Find  the  angles  of  the  rhombus  in  the  last  exercise  (see  page  no,  Exer- 
cise 5). 

23.  A  square  and  a  rhombus  have  the  same  side,  4™.     How  much  smaller  is 
the  latter  if  its  acute  angle  =  60°  ?  if  its  acute  angle  =  30°? 

24:*    A  rectangular  field  2OOm  long  contains  iha.     Find  the  distance  apart  of 
two  opposite  corners. 

25.  The  edge  of  a  cube  —  4m.     Find  the  distance  from  a  corner  in  the  upper 
face  to  the  diagonally-opposite  corner  in  the  lower  face. 

26.  What  method  of  making  a  right  angle  with  a  cord  is  suggested  by  Fig. 


170  GEOMETRY   FOR   BEGINNERS. 

Construct  the  following  figures :  — 

27.  A  triangle,  if  the  area  =  36oicm,  the  altitude  =  42cm. 

28.  An  equilateral  triangle,  if  the  area  =  4OO9cm. 

29.  A  square,  if  the  diagonal  =  1.2™. 

30.  A  rectangle,  if  the  area  =  3001°™,  and  one  side  =  24cm. 

31.  A  rhombus,  if  the  area  =  24O<icm,  and  one  diagonal  =  30°™. 
Construct  a  square,  — 

32.  Three  times  as  large  as  a  given  square. 

33.  Two  and  one-quarter  times  as  large  as  a  given  square. 

34.  Equal  to  half  the  difference  of  two  given  squares. 

35.  Equivalent  to  a  given  equilateral  triangle. 

36.  Equivalent  to  a  given  regular  octagon. 

Given  a  square  with  the  side=  50°™;   compute  the  side  of — 

37.  An  equivalent  equilateral  triangle. 

38.  An  equivalent  regular  pentagon. 

39.  An  equivalent  regular  hexagon. 

40.  An  equivalent  regular  decagon. 
Transform  a  given  triangle,  — 

41.  Into  a  right  triangle. 

42.  Into  an  isosceles  triangle. 

43.  Into  an  isosceles  right  triangle. 

44.  Into  an  equilateral  triangle. 

45.  Into  a  rectangle  with  a  given  side. 

46.  Into  a  square. 
Transform  a  square,  — 

47.  Into  a  parallelogram  with  a  given  angle. 

48.  Into  a  rectangle  with  a  given  side. 

49.  Into  a  rhombus  with  a  given  side. 

50.  Transform  a  given  hexagon  into  a  square. 

51.  What  part  of  a  triangle  is  cut  off  by  a  line  which  bisects  two  of  the  sides? 

52.  Divide  a  triangle  into  three  equal  parts  by  lines  meeting  at  a  given  point 
within  the  triangle,  one  line  to  pass  through  one  corner. 

53*    Find  the  point  within  the  triangle  from  which  lines  drawn  to  the  corners 
will  divide  the  triangle  into  three  equal  parts. 

54.  Divide  a  parallelogram  by  lines  through  a  corner  into  three  equal  parts. 

55.  Divide  a  parallelogram  by  lines  through  a  point  in  one  side  into  three 
equal  parts. 

56.  Divide  a  regular  hexagon  by  lines  through  one  corner  into  six  equal  parts. 

57.  Divide  an  irregular  hexagon  by  a  line  through  one  corner  into  two  equal 
parts. 


$    15^0  CHAPTER   VIII. SIMILAR   FIGURES.  171 

CHAPTER  VIII. 
SIMILAR    FIGURES. 

CONTENTS.  — I.  Proportional  Lines  and  Figures  ($§  156-158).  II.  Similar  Tri- 
angles (§§  159-168).  III.  Problems  and  Applications  (§§  169-176).  IV.  Simi- 
lar Polygons  ($§  177-179). 

J. — Proportional  Lines  and  Figures. 

§  156.   Review  §§  37,  38, 112-114. 

The  lines  a  and  b  (Fig.  150)  have  the  ratio  1:3;  the  lines  c 
and  d  have  the  ratio  2:6=  1:3,  or  the  same  ratio  as  the  lines 
a  and  b.  By  placing  the  sign  of  equality  between  the  equal  ratios 
a :  b  and  c  :  d,  we  obtain  a  proportion,  a  :  b  =  c  :  d,  which  is  read 
a  is  to  b  as  c  is  to  d. 


-2—+  4- 

b 


Fig.  150. 

Definition. — A  PROPORTION  is  an  equation  between  tivo  equal 
ratios. 

The  four  terms  of  a  proportion  need  not  all  be  of  the  same  kind  ; 
it  is  sufficient  if  the  two  terms  of  each  ratio  are  alike  in  kind. 

Two  areas,  for  example,  may  have  the  same  ratio  as  two  lines, 
and  hence  form  a  proportion  with  them.  What  instance  of  this 
occurs  in  §  152  ? 

Again,  two  angles  are  to  each  other  as  the  arcs  described  from 
their  vertices  as  centres  and  included  between  their  sides.  In  this 
proportion,  the  first  ratio  is  that  of  two  angles,  the  second  that  of 
two  lines. 


172  GEOMETRY   FOR    BEGINNERS.  [§     157, 

Let  A,  B,  C,  D  be  any  four  magnitudes  (that  is,  things  that  can 
be  measured) ,  A  and  B  however  being  alike  in  kind,  and  C  and  D 
also  being  alike  in  kind ;  and  let  a  and  b  represent  the  numerical 
values  of  A  and  B  respectively  in  terms  of  a  common  unit,  c  and 
</the  numerical  values  of  C  and  D  respectively  in  terms  of  a  com- 
mon unit ;  then,  if  a  :  b  =  c  :  d,  we  say  that  A>  B,  C,  and  D  are  in 
proportion  or  are  proportionals.  Of  the  four  terms  a,  b,  c,  and  d 
which  express  the  proportion,  a  and  d  are  called  the  extremes,  b 
and  c  the  means.  If  b  =  c,  b  is  called  the  mean  proportional 
between  a  and  d. 

§  157.  A  proportion,  expressed,  as  just  explained,  by  the  num- 
bers which  are  the  numerical  values  of  the  magnitudes  forming  the 
proportion,  is  subject  to  certain  laws,  among  which  the  following 
are  useful  for  our  purposes  :  — 

1 .  If  we  write  the  proportion  a  :  b  =  c  :  d  in  the  form  -=-,, 

and  then  divide  the  equation  i  =  i  by  the  equation  -  =  -,  we 

o       a 

obtain  (Axiom  V.)  -  =  -,  or  b  :  a  =  d\  c. 
a       c 

Law  I.  —  The  truth  of  a  proportion  is  not  affected  if  the  terms 
of  each  ratio  are  transposed. 

2.  If  we  multiply  both  sides  of  the  equation  -  =  -  by  -,  we 

b       d         a 

obtain  (Axiom   IV.),  after  cancelling  common  factors,  -  = -,  or 

d  :  b  =  c  :  a.      If  we  multiply  -  =  -  by  -,  we  obtain  -  =  -,  or 

b       d        c  c       a    . 

a  :  c  =  b  :  d. 

Law  II.  —  The  truth  of  a  proportion  is  not  affected,  if  the  ex- 
tremes are  transposed,  or  if  the  means  are  transposed. 

3.  If  we  multiply  both  sides  of  the  equation  -     =  -    by  the 


§    158.]  CHAPTER   VIII.  -  SIMILAR    FIGURES.  173 

product  b  x  //,  we  obtain,  after  cancelling  common  factors,  a  X  d 
=  b  X  c. 

Law  III.  —  In  every  proportion  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

If  in  a  proportion  the  two  means  and  one  extreme  are  given, 
how  can  the  other  extreme  be  found  ? 

If  in  a  proportion  the  two  extremes  and  one  mean  are  given, 
how  can  the  other  mean  be  found  ? 

Exercises.  —  1.  Verify  the  above  laws  with  the  lengths  of  the  lines  in 
Fig.  150. 

2.   Verify  the  above  laws  in  the  case  of  the  proportion  3:9  =  5:15. 

8.  In  a  proportion,  the  first,  third,  and  fourth  terms  are  12,  25,  200;  find 
the  second  term. 

4.    Find  a  fourth  proportional  to  the  lengths  24™,  50™,  and  8om. 
„  5.    Find  a  length  which  shall  be  to  6om  as  5  :  6. 

6.  What  relation  exists  between  the  side  of  a  square  and  the  sides  of  the 
equivalent  rectangle  ? 

7.  Find  the  mean  proportional  between  9™  and  i6m. 

8.  Show  that  the  mean  proportional  between  a  and  b  is  equal  to  "^Tab. 

§  158.  Let  P  and  Q  denote  the  areas  of  two  squares,  two  par- 
allelograms, or  two  triangles  ;  a  and  b  the  sides  of  the  squares,  or 
the  bases  of  the  parallelograms  or  the  triangles  ;  h  and  k  the  alti- 
tudes of  the  parallelograms  or  of  the  triangles  ;  then,  for  — 

Two  squares,  P=  a\        Q  =  P'}         hence  ^=  -[• 

Two  parallelograms,  P=  aKh,Q  =  bKk]  hence  —  =  a  X  h 


Q      b  X  k 

Two  triangles,  P=  2**,  Q  =  *^-}  hence  1=  **JL 

2  2  Q      b  x  k 

Or,  in  words,  — 

Two  squares  are  to  each  other  as  the  squares  of  their  sides. 

State  in  words  the  results  for  parallelograms  and  for  triangles. 


174  GEOMETRY    FOR   BEGINNERS.  [§    158. 

In  the  cases  of  parallelograms  and  triangles,  if  (i)  h  =  k,  and 
(2)  a  —  b,  the  preceding  equations  become,  — 

(0  ---•  (2)  £-h 

()    Q     *'  Q-J 

That  is,  — 

(  T  )  Two  parallelograms  or  two  triangles,  with  equal  altitudes, 
are  to  each  other  as  their  bases. 

(2)  Two  parallelograms  or  two  triangles,  with  equal  bases,  are 
to  each  other  as  their  altitudes. 

Exercises.  —  1.  The  legs  of  a  right  triangle  are  24™  and  i8m.  In  another 
triangle,  the  base  =  24m,  the  altitude  =  i8m.  Compare  the  areas  of  the  tri- 
angles (that  is,  find  the  ratio  of  the  areas). 

2.  The  bases  of  two  triangles  of  equal  altitude  are  25™  and  3om.     Com- 
pare the  areas. 

3.  If  the  areas  of  two  triangles  of  equal  altitude  are  i6oim  and  2OO<im,  what 
is  the  ratio  of  their  bases?     If  the  base  of  one  —  32™,  find  that  of  the  other. 

4.  If  the  bases  of  two  triangles  of  equal  altitude  are  15"'  and  I9m,  find 
the  base  of  a  triangle  with  the  same  altitude  and  equal  in  area  to  their  sum. 

5.  The  areas  of  two   triangles  with  equal  bases  are  48ocim  and  3001™. 
Compare  their  altitudes.     If  the  altitude  of  the  larger  =  30™,  find  that  of  the 
smaller,  and  also  the  base  of  either. 

6.  Two  triangles  have  equal  bases,  and  the  altitude  of  one  is  ten  times 
that  of  the  other.     Compare  their  areas. 

7.  If  two  triangles  have  equal  areas,  compare  their  altitudes  when  the 
base  of  one  triangle  is  (z.)  twice,  (z'z.)  four  times,  (z'z'z.)  one-half,  (zV.)  three- 
fifths,  that  of  the  other. 

8.  The  altitudes  of  two  equivalent  triangles  are  15°  and  20™.     Compare 
their  bases. 

9.  Solve  Exercises  6,  7,  and  8,  substituting  "  parallelograms  "  for  "  tri- 
angles." 

10.  Compare  a  triangle  with  a  parallelogram  of  the  same  base  and  altitude. 

11.  A  triangle  and  a  parallelogram  have  equal  areas.     Compare  (z.)  their 
bases  if  they  have  equal  altitudes;  (z'z.)  their  altitudes  if  they  have  equal  bases. 

12.  Compare  the  areas  of  a  rhombus  and  the  rectangle  whose  sides  are 
bisected  by  the  corners  of  the  rhombus  (see  Fig.  124}. 

13»  A  square  and  a  rhombus  have  equal  perimeters.  Compare  their  areas 
if  the  altitude  of  the  rhombus  is  two-thirds  its  side. 


§   1 5 9.]  CHAPTER   VIII.  —  SIMILAR   FIGURES.  175 

14.  A  square  and  a  rhombus  have  equal  areas.     Compare  their  perimeters 
if  the  altitude  of  the  rhombus  is  one-half  its  side. 

15.  Compare  the  areas  of  a  square  and  an  equilateral  triangle  having  the 
same  side. 

16.  A  square  and  an  equilateral  triangle  have  equal  areas.     Compare  their 
sides. 

17.  Compare  the  areas  of  a  square,  a  regular  pentagon,  and  a  regular 
hexagon,  all  having  the  same  side. 

II.  —  Similar  Triangles. 

§  159.  Two  triangles  which  have  the  same  shape  and  differ 
only  in  size  are  called  SIMILAR  triangles. 

The  sign  of  similarity  is  ~.     &ABC^>&DEFis  read,  the 
triangle  ABC  is  similar  to  the  triangle  D E F. 

In  order  to  discover  the  chief  properties  possessed  by  similar  tri- 
angles, take  on  one  side  AK  (Fig.  151)  of  an  angle  KAL  any 
number,  say  five,  equal  parts,  AB,  BD,  D  F,  FH,  HK,  and 
through  the  points  B,  D,  F,  H,  K, 
draw  parallel  lines  cutting  A  L  in  the 
points  C,  E,  G,  /,  L ;  then  (§  112) 
AC=CE=EG=  GI=IL. 

The  triangles  ABC,  ADE,  etc., 
differ  in  size,  but  have  the  same 
shape  ;  hence  are  similar. 

Now  compare  the  angles  and  sides 
of  any  two  of  these  triangles,  as  A  DE 
and  AKL.  The  angles  are  equal;  K  Fi  L 

for  the  angle  A  is  common ;  and  of 

the  other  angles,  ADE  =  AKL,  and  AED^ALK  (§  56). 
As  for  the  sides,  it  is  evident  from  the  construction  that  AD  \AK 
=  2:5,  and  likewise  that  A  E  :  AL  =  2  15.  The  same  ratio  also 
exists  between  the  sides  D  E  and  KL ;  for,  if  we  draw  through 
B,  D,  F,  and  PI  lines  parallel  to  AL,  these  lines  will  divide  (by 
§  112)  DE  into  two  equal  parts,  and  KL  into  five  equal  parts, 


176  GEOMETRY    FOR    BEGINNERS.  [§    159. 

and  the  parts  of  DE  will  be  equal  to  those  of  KL  (§  102,  Corol- 
lary i)  ;  therefore,  DE:XL  =  2  15.  Hence,  A D  :  A K  =  A E 
:AL  =  DE\  KL ;  that  is  to  say,  the  sides  opposite  equal  angles 
in  the  triangles  are  proportional. 

The  sides  opposite  equal  angles  are  termed  corresponding  or 
homologous  sides. 

We  should  obtain  like  results  if  the  angle  A  had  any  other  value, 
or  if  the  parallels  BC,  DE,  etc.,  had  any  other  direction,  and  also 
whatever  be  the  length  or  number  of  the  equal  parts  AB,  BD,  etc. 

Therefore,  - 

Theorem  I.  —  Two  similar  triangles  are  mutually  equiangular, 
and  have  their  homologous  sides  proportional. 

Corollary.  —  By  transposing  (§  157,  Law  II.)  the  means  of 
the  proportion  AD:AK=AE:AL,  we  obtain  AD:AE  = 
A  K :  A  L.  This  shows  that,  in  similar  triangles  the  sides  which 
contain  equal  angles  have  the  same  ratio. 

Two  other  theorems  also  follow  from  the  preceding  investiga- 
tion :  — 

Theorem  II.  —  If  a  line  is  drawn  through  two  sides  of  a  tri- 
angle, parallel  to  the  third  side,  a  smaller  triangle  is  formed  which 
is  similar  to  the  given  triangle. 

Theorem  III.  — A  line  drawn  through  two  sides  of  a  triangle, 
parallel  to  the  third  side,  divides  those  two  sides  proportionally. 

Exercises.  —  1 .  The  converse  of  Theorem  III.  is  also  true.  State  it,  and 
illustrate  it  by  means  of  numbers. 

2,  The  three  sides  of  a  triangle  are  84™,  i2Om,  and  ioom.  Through  a  point 
in  the  first  side,  28™  from  the  intersection  of  the  first  and  second  sides,  a  line 
is  drawn  parallel  to  the  third  side.  Find  where  it  will  cut  the  second  side. 
Find,  also,  the  ratio  between  the  triangle  cut  off  and  the  entire  triangle  (§  152). 

3*  If  the  sides  of  a  triangle  are  l6m,  24™,  and  32™,  and  a  line  is  drawn 
from  the  first  side  to  the  second  side,  connecting  points  distant  5m  and  7™, 
respectively,  from  the  intersection  of  the  first  and  second  sides,  will  this  line 
be  parallel  to  the  third  side? 

4.  Draw  two  similar  triangles.  Also,  two  equal  triangles.  Can  two  triangles 
be  equal  without  being  similar?  Can  they  be  similar  without  being  equal? 


§    l6o,]  CHAPTER   VIII.  -  SIMILAR    FIGURES.  177 

§  160.  It  appears  from  the  last  section  that  the  similarity  of 
two  triangles  involves  six  conditions  :  the  equality  of  three  pairs 
of  angles  and  the  equality  of  the  ratios  between  three  pairs  of 
sides.  What  six  conditions  does  the  equality  of  two  triangles 
involve  ? 

We  have  seen  (§§  72-75)  that,  in  general,  from  the  equality  of 
three  parts  in  two  triangles,  we  can  conclude  that  the  triangles  are 
equal  in  all  respects.  We  shall  now  proceed  to  show  that  if  two 
triangles  fulfil  certain  conditions  out  of  the  six  above  mentioned 
as  belonging  to  similar  triangles,  then  these  triangles  must  fulfil 
the  remaining  conditions,  and  hence  must  be  similar  to  each  other, 

§  161.  In  the  triangles  ABC  and  DEF  (Fig.  152)  let  the 
angle  A  ==  D,  B  =  E,  and  therefore 
C^F.  Make  CG  ~  F£>,  and  draw 
GH\\AB.  Then&GJ?C^AD£F 
(I.  Law  of  Equality).  But  A  ABC 
^  A  GHC  (§  159,  Theorem  II.). 
Therefore  A  ABC  ~  A  DEF. 

Hence>-  Fig.  15*. 

Theorem  (I.  Law  of  Similar- 

ity) .  —  Two  triangles  are  similar  if  they  are  mutually  equiangular. 

Exercises.  —  1.  Show  that  two  triangles  are  similar  if  two  angles  of  one' 
are  equal  respectively  to  two  angles  of  the  other. 

2.  What  condition  suffices  to  make  two  isosceles  triangles  similar?  two 
right  triangles? 

8.    Why  are  two  equilateral  triangles  always  similar? 

4.  Construct  three  similar  triangles,  each  having  the  angles  60°  and  80°. 

5.  Construct  upon  a  given  line  A  B  a  triangle  similar  to  a  given  triianglej 


6.  Two  triangles  are  similar  if  their  sides,  taken  two  by  two,  are  parallel. 

7.  Two  triangles  are  s-imilar  if  their  sides,  taken  two  by  two,  are  perpen- 
dicular to  each  other.     Point  out  the  homologous  sides. 

Hints.  —  Revolve  one  of  the  triangles  about  a  corner  through  90°  ;  then  its  sides 
Become  parallel  respectively  to  the  sides  of  the  other  triangle.    See  Exercise  6, 


178  GEOMETRY   FOR    BEGINNERS.  [§    162. 

§  162.  In  the  triangles  ABC  and  DBF  (Fig.  is  2),  let  AC: 
DF=BC\EF,  and  let  also  the  angle  C  =  F.  Make  CG  = 
DF,  and  draw  £.#11  AB\  then  AC:GC=  BC\HC  (§  159, 
Theorem  III.)  Since  the  first  three  terms  of  this  proportion  are 
the  same  as  the  first  three  of  the  proportion  assumed  at  the  start, 
the  fourth  terms  must  also  be  equal;  therefore  CH  =  EF.  Then 
A  GHC  ^  &DEF  (II.  Law  of  Equality).  But  A  ABC  - 
A  GHC  (why  ?):  therefore  A  ABC  ~  A  DEF\  and  hence,  - 

Theorem  (II.  Law  of  Similarity).  —  Two  triangles  arc 
similar  if  they  have  an  angle  equal,  and  the  sides  which  include 
the  angle  proportional. 

§  163.  In  the  triangles  ABC  and  DEF  (Fig.  752),  let  AC 
-.DF=BC:EF,  AOBC,  DF>EF,  and  B  =  E.  Make 
CG  =  DF,  and  draw  GH\\AB;  then  AC-.  GC  =  BC-.HC. 
This  proportion  and  the  proportion  assumed  at  first  have  the  first 
three  terms  equal,  therefore  their  fourth  terms  must  be  equal,  or 
CH  =  EF.  Whence,  A  GHC  ^  DEF  (III.  Law  of  Equality)  . 
ButA^C^  A  GHC,  therefore  kABC^&DEF.  Hence,  - 

Theorem  (III.  Law  of  Similarity).  —  Two  triangles  are 
similar  if  two  sides  of  one  are  proportional  to  two  sides  of  the  other, 
and  the  angles  opposite  the  greater  sides  are  equal. 

§  164.   In  the  triangles  ABC  and  DEF  (Fig.  152],  let 

AC-.DF  =  BC-.EF, 
and  AC\DF=  AB-.DE. 

Make  CG  =  DF,  and  draw  GH  II  AB,  then- 


In  .the  first  and  third  of  these  proportions,  the  first  three  terms 
are  equal,  therefore  the  fourth  terms  must  also  be  equal  ;  that  is, 
CH  —  EF.  For  the  same  reason  it  follows  from  the  second  and 
fourth  proportions  that  GH  =£>E.  Hence,  A  GHC  ^  A  DEF 


§    165.]  CHAPTER  VIII.  —  SIMILAR   FIGURES.  170 

(IV.  Law  of  Equality)  ;  but  A  ABC  ~~  A  GNC,  therefore  &ABC 
^DEF.  That  is,  - 

Theorem  (IV.  Law  of  Similarity).  —  Two  triangles  are 
similar  if  tJicir  sides,  taken  in  order,  are  proportionals. 

Corollary.  —  If  each  side  of  a  triangle  is  two,  three,  four,  etc., 
times  the  homologous  side  of  a  similar  triangle,  then  the  sum  of 
the  sides,  that  is,  the  perimeter,  of  one  triangle  is  two,  three,  four, 
etc.,  times  the  perimeter  of  the  other  triangle  ;  that  is,  — 

In  similar  triangles  the  perimeters  are  to  each  other  as  any  two 
homologous  sides. 

Exercise.  —  Verify  the  truth  of  this  Corollary  by  .two  triangles,  the  sides  of 
one  being  2m,  4™,  5™,  and  those  of  the  other  6m,  12™,  15™. 

§  165.    Let  the  parallels  AC,  DF  (Fig.  153),  be  cut  by  lines 
OA,   OB,   OC,  drawn  through  a  common 
point  O. 

&OAB  ^  &ODE, 

and 


(§  56,  and  I.  Law  of  Similarity).     There- 

fore,-  /     /    _  V 


^  =  :d^Land  —  =—  <*      B> 

OE      DE^      OE     EF  Fig.i53. 


Whence 

The  same  reasoning  might  be  applied  to  any  number  of  parallels 
and  lines  intersecting  them.  Therefore,  — 

Theorem,  —  Lines  drawn  through  a  common  point  divide  par- 
allels into  proportional  parts. 

Exercises.  —  1.  If  one  of  the  parallels  is  divided  into  equal  parts,  how 
will  the  others  be  divided  ? 

2.  If  a  parallel  to  AC  (Fig.  153)  is  drawn  above  O,  cutting  the  lines 
through  O  prolonged,  will  the  theorem  still  hold  true  ?  Can  you  prove  ? 


180 


GEOMETRY    FOR    BEGINNERS. 


[§  i67. 


§  166.   In  the  right  triangle  ABC,  right  angled  at  C  (Fig. 

draw  CD  _L  A  B  ;  it  divides  A  B  into  two 
parts    or    segments.      The    smaller    right 
triangles    A  CD    and    BCD    are    each 
equiangular  with  respect  to  the  triangle 
ABC  (why?);    therefore  they  are  equi- 
angular   with     respect    to    each    other. 
Hence   (I.  Law  of  Similarity),— 
AACD-  therefore  AB  \  AC  =  AC :  AD. 
therefore  AB  :  BC '  =  BC :  BD. 


154- 


Theorems.  —  If  in  a  right  triangle  a  perpendicular  is  drawn 
from  the  vertex  of  the  right  angle  to  the  hypotenuse  :  — 

1.  Either  leg  is  a  mean  proportional  beeween  the  hypotenuse  and 
the  adjacent  segment. 

II.  The  perpendicular  is  a  mean  proportional  between  the  two 
segments  of  the  hypotenuse. 

Exercises.  —  In  a  right  triangle  let  a  and  b  denote  the  legs,  c  the  hypote- 
nuse, h  the  altitude  with  hypotenuse  for  base,  /  and  q  the  segments  of  the 
ihypotenuse.  Find  the  other  quantities,  if,  — 

.1.    a  =  4-5m,  h  =  3.6™.  3.    a  =  9.6™,  /  —  o.784m. 

2.  /&  =  o.8m,/  =  i.5m.  4.  c= 


167.  Let  (Fig.  155)  kABC^&DEF,  and  draw  the  alti- 
tudes CG  and  FH  corresponding  to 
two  homologous  sides  AB  and  DE 
taken  as  bases. 

By  hypothesis,  AB\DE  =  AC; 
DF.     The  right  triangles  ACG  and 
DFH   are    similar    (why?),    there- 
fore A  C  :  DF  =  CG  :  FH.     Hence 
AB-.DE  =  CG'.FH.     Thatis,- 
Theorem.  —  In  two  similar  triangles,  any  two  homologous  sidet 
taken  as  bases  are  to  each  other  ax  the  corresponding  altitudes*   • 


§   1  68.]  CHAPTER  VIII.  —  SIMILAR   FIGURES.  181 

§  168.   Let  P  and  Q  denote  the  areas,  a  and  b  the  bases,  h  and 
k  the  corresponding  altitudes  of  two  similar  triangles. 

Then  (,  158), 
But  (|  167) 

If  for  the  fac 
the  equation  becomes,  — 


If  for  the  factor  -  in  the  first  equation  we  substitute  its  equal  -, 
k  b 


=±    y    f   =    « 

<2     J     S     £2' 

Since  «  and  /£  may  be  any  two  homologous  sides  taken  as  bases, 
therefore,  — 

Theorem.  —  Similar  triangles  are  to  each  other  as  the  squares 
of  their  homologous  sides. 

Exercises.  —  1.  If  the  homologous  sides  of  one  triangle  are  each  one-fifth 
those  of  a  similar  triangle,  compare  the  areas  of  the  triangles. 

2.  The  areas  of  two  similar  triangles  are  4OO<im  and  1  296im  ;  find  the 
ratio  of  two  homologous  sides. 

8*  Two  triangular  gardens  are  similar  in  shape.  What  is  the  easiest  way 
(z.)  to  compare  their  areas  ?  (zY.)  to  compute  both  their  areas  ? 

4.  A  field  has  the  shape  of  an  equilateral  triangle.     From  one  corner  I 
wish  to  cut  off  a  similar  triangle  equal  in  area  to  one-sixteenth  of  the  whole  ; 
how  must  the  line  of  division  be  run  ? 

Draw  a  triangle  ;   then  construct  a  similar  triangle  equal  in  area  to  — 

5.  4  times  that  of  the  first  triangle. 

6.  2  times  that  of  the  first  triangle. 
7»    \  that  of  the  first  triangle. 

8.   T9g-  that  of  the  first  triangle. 

•    9.    Can  you  prove  the  theorem  of  this  section  by  a  method  suggested  by 
§  152,  Case  2  ? 

Hints.  —  Let  a  line  cut  i,  £,  |,  or  any  other  fraction,  from  each  of  two  sides  of 
a  triangle;  and  use  §  159,  Exercise  i,  and  Theorem  II. 


182 


GEOMETRY   FOR   BEGINNERS. 


[§    I69. 


IIL— Problems  and  Applications. 

§  169.   Problem.  —  To  find  the  fourth  proportional  to  three 
given  lines  a,  b,  c  {Fig.  156). 

Construction.  —  Make  any  angle  O ;  on  its  sides  take 


^       1      I 

O                                DC 
Fig.  156. 

4.    b  >  a,  and  c  >  b. 
5.    b  >  a,  and  c  >  a. 

6. 

7. 

and  draw  BD  II  AC.  OD 
is  the  fourth  proportional  re- 
quired. 

Proof.  —  Apply    §    159, 
Theorem  III. 

Exercises.  —  Give    the    con- 
struction for  the  following  cases  : 

1.  b<a,  and  c> a. 

2.  b<a,  and  c  =  a. 

3.  b  <  c,  and  c  =.  b. 
b> a,  and  c  —  a. 

7,    b>a,  and  c  =  b. 


§  170.   Problem.  —  71?  find  a  mean  proportional  to  two  given 
lines  a  and  b  (Fig.  157} . 

§   169  furnishes  one  solution  (see  Exercises  3  and  7).     The 

_2 more    common    solution    is   as   fol- 

_£ lows  :  — 

Construction.  —  Draw   a    straight 
™  Xf>x  line,  and  take  A  B  =  a,  B  C  =  b.    At 

\\  B  erect  a  perpendicular.     Bisect  A  C 

\\         in  O,  and  with  a  radius  equal  to  O  A 
\\        describe  an  arc  cutting  the  perpen- 
~~0      B   '        c      dicular  in  D.     BD  is  the  mean  pro- 
portional required. 

Proof.  —  Prove,  as  in   §   149,  that  ADC  =  90°;   then  apply 
166. 


,*'' 


§  1 7i.] 


CHAPTER   VIII.  —  SIMILAR   FIGURES. 


183 


§  171.    Problem.  —  To  reduce  or  enlarge  given  lines  a,  b,  c, 
etc.  (Fig.  158) ,  in  a  given  ratio. 

Construction.  —  Let  the  ratio  be  3  :  4  ;  that  is,  let  it  be  required 
to  reduce  the  lines  each  to 
three    fourths     its     present       ~~~ 
length. 

Draw  a  straight  line  and      ' 
take  on  it  four  equal  parts, 
of  which  OA  contains  four, 
OB   three.     At  A   and   B 
erect  perpendiculars ;  make 


etc. 
etc. 


Join  OC,  OD,  OE, 
BF,  BG,   BH,  etc., 


O  B  A 

Fig.  158. 
are  the  reduced  lengths  required. 

Proof.  —  &OBF<^  AOAC,  AO3G  ^  AOAD,  etc.  (why?). 
Then  apply  §  159,  Theorem  I. 

Exercises.  —  1.    Draw  four  lines  and  reduce  them  in  the  ratio  3:  5. 
2.    Draw  four  lines  and  enlarge  them  in  the  ratio  5  :  2. 

§  172.    Problem.  —  To  divide  a  given  line  into  equal  parts. 

One  solution  of  this  problem  has  been  given  (§  113) ;  but,  since 
that  solution  requires  the  construction  of  several  parallels,  it  is 
tedious  and  liable  to  lead  to  errors. 

The  following  construction  is  more 
simple  :  — 

Construction.  —  Let  it  be  required 
to  divide  MN  (Fig.  159)  into  five 
equal  parts.  Draw  any  line  MO 
through  M-,  then  through  a  point 
P  of  this  line  draw  a  line  PQ  II 
MN,  and  take  on  this  parallel  any 
five  equal  parts  ending  at  a  point  Q.  pig. 


M 


N 


184 


GEOMETRY  FOR  BEGINNERS. 


[Si7> 


Through  TV7 arid  Q  draw  a  line  cutting  MO  at  O.    Lastly,  draw  lines 
through    O  and  the  points  of  division  of  PQ.     These  lines  will 
divide  MN  into  five  equal  parts. 
Proof.  —  Apply  §  165. 

Exercises.  —  1.    Draw  a  line  and  divide  it  into  eight  equal  parts. 
2.    Divide  the  line  AB  {Fig.  166}  into  ten  equal  parts. 
Solution.  —  Here  the  parts  are  so  short  that  if  obtained  in  the  way  above  de- 
scribed they  would  be  indistinct.     In  this  case  pro- 

C, ,  -D     ceed  thus  :  Erect  at  A  and  B  perpendiculars,  lay  off 

upon  each  ten  equal  parts,  and  through  the  points 
of  division  draw  lines  parallel  to  A  B.  If  now  we 
draw  the  diagonal  AD,  then  the  distances  from  the 
points  marked  i,  2,  3,  4,  etc.,  to  the  line  BD  will  be 
equal  to  fa,  fa,  T3o»  To>  etc«.  of.  the  line  A  B.  The 
proof  is  left  to  the  learner. 

3.  When  (in  making  maps,  plans,  etc.)  many 
lines  have  to  be  very  much  reduced  in  length,  it 
becomes  often  convenient  to  employ  scales  of 
equal  parts,  known  among  surveyors  as  PLAT- 
TING SCALES.  Explain  how  such  a  scale  (shown 
in  Fig.  161},  containing  one  thousand  parts,  di- 
vided decimally,  is  constructed  and  used. 

Solution.  —  Draw  a  straight  line,  and  take  on  it 
ten  equal  parts,  A [B  =  B  C=  CD,  etc.,1  each  of  these 
lengths  representing  one  hundred  units  of  length 
(meters).  Divide  AB  into  ten  equal  parts,  each 
Pip  160.  °  part  representing  therefore  ten  units  of  length.  Last- 

ly, to  represent  single  units  of  length,  draw  above 
AB  ten  equidistant  lines  parallel  to  A  Bt  erect  perpendiculars  at  the  points  A,  B,  C, 


9  8  7  6 


c 

Fig.  161. 


1  Only  a  part  of  the  scale  can  be  shown  in  the  figure. 


§   1 73-]  CHAPTER   VIII. SIMILAR    FIGURES.  185 

etc.,  and,  after  having  subdivided  FO  into  ten  equal  parts,  join  the  points  of  division 
of  FO  with  those  of  A  D  by  lines  drawn  obliquely,  as  shown  in  the  figure. 

The  perpendicular  OB,  divided  into  ten  equal  parts  by  the  parallels  is  num- 
bered from  L  to  10,  beginning  at  O\  then  it  follows  (see  Exercise  2)  that  the  dis- 
tance from  i  to  the  first  oblique  line  measured  on  the  parallel  =  i  unit  of  length, 
the  distance  from  i  to  the  second  oblique  line  =  II  units,  etc. ;  and  the  distance 
from  2  to  the  first  oblique  line  =  2  units,  the  distance  to  the  second  oblique  line  = 
12  units,  etc.;  and  similarly  for  the  other  points  of  division  of  OB. 

The  point  marked  O  is  the  zero  of  the  scale ;  hundreds  are  read  off  to  the  right, 
tens  to  the  left,  and  units  on  the  vertical  line. 

From  this  explanation  it  is  easy  to  see  that  a  length  of  348  meters  measured  on 
the  ground  is  represented  upon  the  scale  by  the  line  mn  ;  also,  that  the  length  p  q 
corresponds  to  a  distance  of  216  meters,  the  length  rs  to  183  meters,  etc. 

4.  Construct,  with  the  aid  of  the  scale  in  Fig.  161,  a  triangle  whose  sides 
are  137™,  i6om,  and  225"*. 

5.  Draw  a  triangle,  and  determine  the  lengths  of  its  sides  by  means  of  the 
scale  in  Fig.  161. 

6.  Construct  a  scale  of  one  hundred  parts,  divided  decimally. 

§  173.  The  measurement  of  the  distance  of  an  inaccessible 
object  is  one  of  the  most  interesting  applications  of  the  laws  of 
Geometry.  It  gives  us  a  good  idea  of  the  nature  of  those  methods 
which  have  enabled  astronomers  to  measure  the  distance  from  the 
earth  to  the  moon,  to  the  sun,  and  even  to  some  of  the  fixed  stars. 

We  have  already  seen  (§§  95-98)  how  distances  may  be  meas- 
ured indirectly  by  means  of  equal  triangles.  Now  it  is  clear  that, 
if  equal  triangles  are  employed,  the  greater  the  distance  to  be 
measured  the  larger  the  triangle  which  must  be  constructed ;  so 
that,  when  the  distance  is  considerable,  the  construction  of  the 
triangle  becomes  very  inconvenient,  and  in  many  cases  (for  in- 
stance, the  distance  from  the  earth  to  the  moon)  utterly  impos- 
sible. In  such  cases  other  methods  must  be  found ;  and  these 
methods  are  supplied  by  the  laws  of  similar  triangles. 

In  fact,  the  second  method  given  for  solving  the  problems  of 
§  95  and  §  98  is  based  upon  the  properties  of  similar  triangles. 
In  §  95,  Method  II.,  we  construct  on  paper  to  a  reduced  scale  a 
fac-simile  of  the  actual  lines  on  the  ground.  What  is  \kasfac-simik 
but  a  small  triangle  similar  to  the  much  larger  one  which  contains 
the  distance  sought?  In  other  words,  the  triangles  ABC  and  abc 


186  GEOMETRY   FOR   BEGINNERS.  [§    174. 

(Fig.  88)  are  similar;  hence  ac  :  ab  =AC :  A  B,  a  proportion  which, 
by  making  (as  on  page  105)  ac  =  i6cnv,  ab  =  24cm,  AC  =  8oom, 

becomes  16:24  =  800  :  AB ;  whence  AB  =  24  X  8oQ  =  i2oora. 

16 

Representing  lines  to  scale  on  paper,  all  angles  remaining  un- 
changed, always  gives  a  figure  similar  to  that  formed  by  the  unre- 
duced lines,  and  is  termed  PLATTING  TO  SCALE. 

Some  examples  will  now  be  given  of  the  indirect  measurement 
of  distances  by  means  of  similar  triangles. 

§  174.  Problem.  —  To  measure 
the  distance  of  two  points  A,  X,  sepa- 
rated by  a  river. 

At  the  point  A  (Fig.  162),  sup- 
posed accessible,  erect  (with  a  sur- 
veyor's cross  or  other  means)  a  line 
_L  AX,  and  measure  its  length  to 
any  point  B.  Suppose,  for  exam- 
ple, that  AB  —  i20m;  take  BC  = 
iom,  and  at  C  erect  a  second  per- 
pendicular CD,  meeting  the  line  B X  at  D.  Measure  CD ;  let  its 
length  be  i8m.  Then  it  follows  that,  in  all  right  triangles  similar 
to  BCD  —  that  is  to  say,  in  all  right  triangles  having  an  acute 
angle  equal  to  DBC — the  ratio  of  the  legs  must  be  18 :  IO.1 

Among  these  similar  right  triangles  is  the  triangle  ABX;  there- 
fore, AX-.AB  =  18:  10;  whence,  AX=^2i6m. 

Exercises.  —  1.  In  Fig.  162,  BC  is  made  equal  to  iom  solely  for  con- 
venience of  computation.  If  B  C=  15™,  what  would  D  C  be  found  to  be  ? 

2.  If  A  is  on  the  edge  of  the  river,  can  you  find  A  X  by  this  method  ? 
What  change  in  the  position  of  the  triangle  B  CD  is  necessary  ?     Give  the 
construction  for  this  case. 

3.  How  can  AX  be  found  by  means  of  two  equal  right  triangles  ? 

4.  How  can  A  X  be  found  by  means  of  an  isosceles  right  triangle  ? 

1  In  Trigonometry,  18 : 10,  or  {£,  is  called  the  TANGENT  of  the  angle  DBC. 


§  I75-] 


CHAPTER   VIII. SIMILAR   FIGURES. 


187 


§  175.    Problem.  —  To  find  the  distance  between  two  inac- 
cessible points,  X,  Y. 

Case  1  (Fig.  i6j).  —  Choose  an  accessible  point,  A,  and  meas- 
ure the  distances  AX,  AY,  by 
the  preceding  problem.  Upon 
AX  measure  a  length  Ax  equal 
to  any  fractional  part  (as  TV,  5^, 
etc.)  of  AX,  and  upon  A  Y  meas- 
ure Ay  equal  to  the  same  frac- 


tional part  of  A  Y.  Lastly,  meas- 
ure xy,  which  is  parallel  to  X  Y 
(why  ?) .  Then  the  distance  X  Y  can  be  found  from  the 


Fig.  163. 


If,  for  example,  Ax  =  -fa  AX,  then  XY  =  10  xy. 

Case  2.  —  Where  the  line  X  Y  prolonged  is  accessible  (fig.  164)  . 
Erect  at  M.  an  accessible  point,  a  line  J_  MX,  and  mark  the  points 
N  where  this  line  intersects  the  perpendicular  let  fall  from  Y,  and 
P,  where  it  cuts  XY  prolonged.  Measure  MP  and  NP;  let,  for 
example,  MP  =  360™,  NP  =  200™. 
Take  PO  =  iom  ;  at  O  erect  OQ  _L 
PM,  and  measure  the  hypotenuse 
PQ  of  the  right  triangle  POQ.  Sup- 
pose, for  example,  that  PQ  =  24™  ; 
then,  in  all  right  triangles  having  an 
acute  angle  equal  to  OPQ  the  ratio 
of  the  leg  corresponding  to  OP  to  the 
hypotenuse  is  10  :  24.* 

PMX  and  PNY  are  two  such  tri- 
angle?    therefore,  — 


N 
Fig.  164. 


M 


10  :  24  =  PM,  or  360™ 

10  :  24  =  PN,  or  200"= 

and  XY=  PX  -  PY  =  384* 


PX,  whence  PX  =  864m  ; 
PY,  whence  PY  =  48om ; 


1  In  Trigonometry,  the  ratio  10:  2.4,  or  j5^,  is  called  the  COSINE  of  the  angle  OPQ. 


iss 


GEOMETRY    FOR    BEGINNERS. 


[§ 


Exercises.  —  1.    What  is  the  advantage  in  Case  2  of  making  P  0  =  iom 
rather  than  any  other  number  ? 

2.  Explain  how,  with  the  aid  of  an  instrument  for  measuring  angles,  the 
distance  X Fin  Fig.  163  may  be  found  by  platting  a  similar  triangle  on  paper. 
By  this  method  how  many  lines  must  be  measured  on  the  ground  ?     How 
many  by  the  other  method  ? 

3.  Given  (Fig.  163}  A  X  =  6oom,  A  Y=  720™,  Ax  =  ±5  AX,  Ay  =  -fa  AY, 
xy-2i.iSm,  XAY=  36°.     Find  XY  (z.)  by  a  direct  proportion;    («.)  by 
platting  a  similar  triangle  to  scale   on  paper.     How  nearly  do   the  results 
agree  ?     Why  do  they  not  exactly  agree  ? 

4.  Fig.  zdj  illustrates  another  method  of  measuring  an  inaccessible  line 
XY,  —  a  method  by  which  only  one  line  AB  has  to  be  measured  on  the 
ground.     Can  you  explain  the  method  ? 


Fig.  165. 


§  176.  Problem.  —  To  measure  the  vertical  height  of  an  ob- 
ject (a  tower,  tree,  church- spire,  etc.) 

1st  Method.  —  By  means  of  a  shadow. 

Suppose  that  we  wish  to  find  the  height  A  X  of  a  tree  (Fig. 
166)  which  casts  the  shadow  'A  B  on  the  ground.  Fix  a  staff  ver- 
tically in  the  ground  near  the  tree,  and  measure  its  height  mn. 
Measure,  also,  the  lengths  of  the  shadows  n  o  cast  by  the  staff,  and 
AB  cast  by  the  tree.  Now  the  ray  of  light  XB  is  parallel  to  the 
ray  mo :  whence  it  follows  that  ABX  and  mno  are  similar  trian- 


1 76.] 


CHAPTER   VIll. SIMILAR    FIGURES. 


189 


gles  (why  ?) .     In  other  words,  the  shadows  cast  at  the  same 
by  different  objects  are  proportional 
to  the  heights  of  the  objects. 

Therefore  no  :  AB  =  m n  :  A X. 
If  no  =  0.9™,  AB  =  24. 9m,  mn  = 
i.i7m,  then, — 

0.9    :  24.9  =1.17    :  AX, 
whence       AX  =32. 37™. 

NOTE. —  We  may  also  reason  thus  :  if  9Ocm 
of  shadow  correspond  to  H7cin  of  height,  then 
lcm  of  shadow  corresponds  to  -U_7  =  1.3^1  of 
height ;  therefore  249ocm  of  shadow  will  cor- 
respond to  i.3cm  X  2490  =  3237°™  =  32.37m. 

2d  Method.  —  With  the  aid  of  an 
instrument  for  measuring  angles. 

Let,  for  example,  the  height  of 
the  tower  (Fig.  167)  be  required.  Place  the  instrument  at  a  point 
B,  some  distance  from  the  foot  of  the  tower,  and  measure  the 
angle  abX,  or  angle  of  elevation,  as  it  is  termed.  Measure  AB, 
the  distance  from  the  instrument  to  the  foot  of  the  tower.  Then 
construct  on  paper  a  right  triangle  similar  to  abX,  and  the  side 
of  this  triangle  homologous  to  a  X  will  give  to  the  scale  employed 
the  height  aX,  to  which  we  must  add  A  a,  in  order  to  obtain  the 
entire  height. 

pnnnnc 


Fig.  166. 


H 


4- 

"  Fig.  167.  Fig.  168. 

3d  Method.  —  When  the  base  of  the  object  is  inaccessible,  choose 


190  GEOMETRY   FOR   BEGINNERS.  [§    177. 

a  convenient  place  near  the  object,  and  measure  a  base  line  AB 
(Fig.  1 68).  Then  measure  the  angles  at  the  base,  B  and  B  A  X, 
and  also  the  angle  of  elevation  of  the  object  XAO,  and  construct 
on  paper  a  triangle  similar  to  XB  A.  This  will  give  to  scale  the 
length  A  X.  Then,  knowing  A  X  and  the  acute  angle  X A  O,  con- 
struct on  paper  the  right  triangle  OA  X,  which  will  give  to  scale 
the  height  OX  required. 

Here  the  unknown  height  OX  depends  on  the  construction  of 
two  similar  triangles,  the  first  similar  to  the  triangle  ABX,  the 
second  to  the  right  triangle  OAX.  If  the  scale  of  reduction  in 
both  cases  is  imm  to  im,  the  number  of  millimeters  contained  in 
the  line  on  paper  which  represents  OX  will  be  the  same  as  the 
number  of  meters  in  OX. 

Exercises.  —  1.  A  monument  casts  a  shadow  of  44™  at  the  same  time  that: 
a  rod  2m  high  casts  a  shadow  of  i.33m  ;  find  the  height  of  the  monument. 

2»  In  the  2d  Method,  why  is  the  construction  of  a  triangle  on  paper 
avoided  by  placing  the  instrument  so  that  the  angle  of  elevation  shall  be 
equal  to  45°  ? 

3.  What  instruments  are  required  in  the  3d  Method  ? 

4.  Find  the  height  of  the  building  (Fig.  168}  if  A  B  =  250™,  angle  B  = 
25°  30',  angle  XA  B  =  104°,  and  angle  XA  0  =  41°  30'. 

IV.— Similar  Polygons. 

§  177.  Two  polygons  are  said  to  be  SIMILAR  if  they  have  the 
same  shape. 

In  order  to  see  what  are  the  properties  of  similar  polygons, 
divide  the  polygon  A  B  CD  E  (Fig.  169)  into  triangles  by  drawing 
the  diagonals  AC,  AD ;  then  begin  at  any  point  M  of  AB,  and 
make  a  new  polygon  by  drawing  MN  II  J3C,  NO  II  CD,  OP  II  DE. 
The  new  polygon  A  MNO  P  is  smaller  than  the  original  polygon, 
but  has  the  same  shape,  and  is  therefore  similar  to  it. 

By  the  same  process,  we  can  make  a  polygon  AQRST  larger 
than  the  polygon  ABCDE,  and  similar  to  it. 


§    I  77-]  CHAPTER    VIII.  —  SIMILAR    FIGURES.  191 

Now  the  triangles  AMN,  ABC,  AQR,  are  similar  (why?);  so 
are  the  triangles  AON,  ADC,ARS\ 
and  lastly,  the  triangles  A  OP,  ADE, 
A  S  T.     Whence  it  follows  that,  — 

Theorem  I.  —  Similar  polygons 
are  composed  of  the  same  number  of 
triangles  similar  to  each  other  and 
similarly  placed. 

From  this  it  follows,  that  similar 
polygons  are  mutually  equiangular. 

Why,    for    example,    is    the     angle         A  M      B      o 

AMN  equal   to  the  angle  ABC?  Fig.  169. 

the  angle  MNO  equal  to  the  angle  BCD? 

Compare  now  the  sides  in  any  two  of  the  polygons,  say  the  two 
smaller  ones.     From  the  similar  triangles  AMN  and  ABC,  — 

AM:AN  =  MN-.BC, 
and  also,  AN:  AC  =  MN:  BC-, 

and  from  the  similar  triangles  A  NO  and  AC  D, 

AN-.  AC  =  NO:  CD. 
From  the  second  and  third  proportions,  it  follows  (Axiom  I.)  that 


In  this  way,  whatever  be  the  number  of  sides,  it  may  be  shown 
that  the  sides  similarly  placed  are  always  proportional. 

Theorem  II.  —  Similar  polygons  are  mutually  equiangular, 
and  have  their  homologous  sides  proportional. 

Corollaries.  —  I  .  Two  regular  polygons  of  the  same  number 
of  sides  are  similar.  Why? 

2.  If  each  side  of  a  polygon  is  2,  3,  4,  etc.,  times  as  great  as 
the  homologous  side  of  a  similar  polygon,  then  the  sum  of  the  sides, 
that  is,  the  perimeter  of  the  first  polygon,  must  be  2,  3,  4,  etc., 
times  as  great  as  the  sum  of  the  sides,  or  the  perimeter,  of  the  second 
polygon.  That  is  to  say,  the  perimeters  of  similar  polygons  are  to 
each  other  as  any  two  homologous  sides. 


192  GEOMETRY    FOR    BEGINNERS.  [§    178. 

§  178.  If  two  similar  polygons,  the  sides  of  one  of  which  are 
double  those  of  the  other,  are  divided  into  triangles,  by  drawing 
diagonals  from  the  vertices  of  two  equal  angles,  then  each  triangle 
in  the  first  polygon  is  (by  §  164)  four  times  as  great  as  the  corre- 
sponding triangle  on  the  second  polygon ;  hence  the  sum  of  all 
the  triangles  in  the  first  polygon  —  in  other  words,  the  area  of  the 
first  polygon  —  is  four  times  as  great  as  the  sum  of  all  the  triangles, 
or  the  area,  of  the  second  polygon.  Hence  the  areas  of  the  two 
polygons  are  to  each  other  as  4  :  i.  But  this  is  the  ratio  of  the 
squares  of  any  two  homologous  sides.  A  like  result  would  follow 
if  the  sides  had  any  other  ratio  than  2:1.  Hence,  — - 

Theorem.  —  Similar  polygons  are  to  each  other  as  the  squares 
of  their  homologous  sides. 

Therefore,  if  we  have  a  polygonal  figure  on  the  ground,  and 
construct  a  similar  figure  on  paper  with  its  sides  reduced  to  £,  ^, 
|-,  TV,  etc.,  of  their  lengths,  the  area  of  the  figure  on  the  paper  will 
be  i>  FJ  TB>  T<y<i>  etc'>  the  area  of  the  larger  and  similar  figure  on 
the  ground. 

Exercises.  —  1.  Compare  the  areas  of  two  similar  polygons  with  their 
perimeters  (see  §  177,  Corollary  2). 

2*  Prove  that  the  areas  of  regular  polygons  of  the  same  number  of  sides 
are  to  each  other,  (z.)  as  the  squares  of  their  sides ;  (zY.)  as  the  squares  of 
their  perimeters. 

3.  If  in  two  hexagonal  parks  a  side  of  one  is  four  times  a  side  of  the 
other,  how  much  larger  is  one  park  than  the  other  ?  Find,  also,  the  ratio 
of  their  perimeters. 

§  179.  Problem.  —  To  construct  a  polygon  similar  to  a  given 
polygon. 

There  are  various  methods  of  solving  this  problem. 

1st  Method.  —  When  one  side  F  G  of  the  required  polygon  is 
given. 

Solution.  —  By  means  of  §  177,  Theorem  I.  Divide  the  given 
polygon  ABODE  (Fig.  170)  into  triangles  by  diagonals;  then 


§  1 79-] 


CHAPTER   VIII. SIMILAR    FIGURES. 


193 


construct  upon  the  given  length  FG,  homologous  to  AB,  a  series 
of  triangles  similar  to  those  of  the  given  polygon  respectively,  and 
similarly  placed.  The  poly- 
gon FGHKL  is  the 
polygon  required. 

Give  the  construction  in 
full. 

2(1  Method.  —  One   side 
F  G  being  given,  as  before. 

Solution.  —  On  one  side 
AB  of  the  given  polygon 
ABCDE  (Fig.  777)  take 
AAf=FG,  construct  the 
polygon  AMNOP^  ABCDE  (how  is  this  done?),  and  then 
construct  (by  §  117)  upon  the  side  FG  the  polygon  FGHKL 
&  the  polygon  AM  NO  P. 

Give  the  construction  in  full. 


H 


Fig.  170. 


F 


3d  Method.  —  When  the  homologous  sides  of  the  two  polygons 
are  to  be  to  each  other  in  a  given  ratio. 

Solution.  —  Choose  any  point  O,  either  within  or  without  the 
given  polygon  (see  Figs.  172  and  //j),  and  join  it  to  all  the  cor- 
ners by  straight  lines.  Divide  one  of  these  lines  at  a  point  Af,  so 
that  O M\  OA  —  the  given  ratio,  and  then  draw  MN  II  AB,  NP 
II  BC,  etc.  MNPQR  is  the  required  polygon.. 


194 


GEOMETRY    FOR    BEGINNERS. 


[§    179. 


In  Fig.  172,  OM\  OA  =  i  :  3  ;  in  Fig.  173.  OM\  OA  =  3  :  5. 
Give  the  construction  and  proof  for  both  cases. 


Fig.  172. 


Fig.  173. 


Exercises.  —  1.    Given  a  hexagon;  construct  a  similar  but  larger  hexagon. 

2.  Given  an  octagon;   construct  a  similar  but  smaller  octagon. 

3.  Upon  the  lengths  2Ocm  and  30°™  as  homologous  sides  construct  two 
similar  pentagons. 

4.  Draw  a  polygon,  and  then  construct  another  whose  sides  shall  be, — 

(z.)      one-half  those  of  the  first  polygon  ; 
(it.}    two-thirds  those  of  the  first  polygon  ; 
(tit.')  twice  those  of  the  first  polygon  ; 
(zz/.)    three  and  one-half  times  those  of  the  first  polygon. 
5*    Draw  a  polygon,  and  then  construct  a  similar  polygon, — 
(z.)      one-ninth  as  large  as  the  first  (§  174) ; 
(z'z.)     nine  twenty-fifths  as  large  as  the  first ; 
(z'z'z.)  four  times  as  large  as  the  first ; 
(zV.)    twice  as  large  as  the  first. 

6.  Construct  two  similar  pentagons  which  shall  be  to  each  other  as  4:  9. 

7.  Construct  two  similar  regular  pentagons,  one  having  three  times  the 
perimeter  of  the  other.     What  is  the  ratio  of  their  areas? 

8.  Construct  two  regular  hexagons  about  the  same  point  as  centre,  making 
the  less  radius  of  one  =  l\  times  that  of  the  other. 

9.  Construct  two  regular  octagons,  making  the  less  radius  of  one  =  f  that 
of  the  other.     What  is  the  ratio  of  their  perimeters  ?  of  their  areas  ? 


CHAPTER   VIII.  —  REVIEW.  195 

REVIEW   OF   CHAPTER  VIII. 
SYNOPSIS. 

1.  A  proportion  is  an  equation  between  two  equal  ratios. 

2.  The  truth  of  a  proportion  is  not  affected  if  the  terms  of  both  ratios  are 
transposed. 

3.  The  truth  of  a  proportion  is  not  affected  if  the  extremes  are  transposed, 
or  if  the  means  are  transposed. 

4.  In  every  proportion  the  product  of  the  extremes  is  equal  to  the  product 
of  the  means. 

5.  Two  squares  are  to  each  other  as  the  squares  of  their  sides;   two  paral- 
lelograms, or  two  triangles,  as  the  products  of  their  bases  by  their  alti- 
tudes. 

6.  If  two  parallelograms,  or  two  triangles,  have  equal  bases,  they  are  to 
each  other  as  their  altitudes ;    if  they  have  equal  altitudes,  they  are  to 
each  other  as  their  bases. 

7.  Similar  triangles  are  triangles  that  have  the  same  shape. 

8.  Similar  triangles  are  mutually  equiangular,  and  have  their  homologous 
sides  proportional. 

9.  In  two  similar  triangles,  the  sides  that  include  equal  angles  have  the 
same  ratio. 

10.  A  line  drawn  through  two  sides  of  a  triangle  parallel  to  the  third  side 
cuts  off  a  smaller  triangle  similar  to  the  entire  triangle. 

11.  A  line  drawn  through  two  sides  of  a  triangle  parallel  to  the  third  side 
divides  those  sides  proportionally. 

12.  I.    Law  of  Similarity.  —  Two  triangles  are  similar  if  they  are  mutually 
equiangular. 

13.  II.    Law  of  Similarity. — Two  triangles  are  similar  if  they  have  an 
angle  equal,  and  the  sides  including  this  angle  proportional. 

14.  III.    Law  of  Similarity.  —  Two  triangles  are  similar  if  two  sides  of  one 
are  proportional  to  two  sides  of  the  other,  and  the  angles  opposite  the 
greater  sides  are  equal. 

15.  IV.    Law  of  Similarity.  — Two  triangles  are  similar  if  their  sides,  taken 
in  order,  are  proportional. 

16.  Lines  through  a  common  point  divide  parallels  proportionally. 

17.  If  in  a  right  triangle  a  perpendicular  is  drawn  from  the  vertex,  of  the  right 
angle  to  the  hypotenuse,  — 


1.96  GEOMETRY    FOR    BEGINNERS. 

(1)  each  leg  is  a  mean  proportional  between  the  hypotenuse  and  the 
adjacent  segment ; 

(2)  the  perpendicular  is  a  mean  proportional  between  the  segments  of 
the  hypotenuse. 

18.  The  homologous  sides  of  similar  triangles  are  to  each  other  (i)  as  the 
corresponding  altitudes,  (2)  as  the  perimeters. 

19.  Similar  triangles  are  to  each  other  as  the  squares  of  their  homologous 
sides. 

20.  Problem.  —  To  find  the  fourth  proportional  to  three  given  lines. 

21.  Problem.  —  To  find  the  mean  proportional  between  two  given  lines. 

22.  Problem.  —  To  reduce  or  enlarge  given  lines  in  a  given  ratio. 

23.  Problem.  —  To  divide  a  given  line  into  equal  parts.     (Two  ways.) 

24.  Problem.  —  To  measure  the  distance  from  an  accessible  to  an  inaccessi- 
ble point. 

25.  Problem.  —  To  measure   the   distance  between  two  inaccessible  points. 
(Two  cases.) 

26.  Problem.  —  To  measure  the  height  of  an  object.     (Three  methods.) 

27.  Similar  polygons  are  polygons  that  have  the  same  shape. 

28.  Similar  polygons  are  composed  of  the  same  number  of  similar  triangles 
similarly  placed. 

29.  Similar  polygons  are  mutually  equiangular,  and  have  their  homologous 
sides  proportional. 

30.  Regular  polygons  of  the  same  number  of  sides  are  similar. 

31.  The  perimeters  of  similar  polygons  are  to  each  other  as  any  two  homo- 
logous  sides. 

32.  Similar  polygons  are  to  each  other  as  the  squares  of  their  homologous 
sides. 

33.  Problem.  — To  construct  a  polygon  similar  to  a  given  polygon.     (Three 
methods.) 

EXERCISES. 

1.  A  square  and  a  rhombus  have   equal  areas  :    find  the  ratio  of  their 
perimeters,  if  the   altitude   of  the   rhombus  is   one-fourth  that  of  the 
square. 

2.  What  single  condition  will  make  two  right  triangles  similar  ?    also  two 
isosceles  triangles  ? 

3.  Given  a  triangle,  two  of  whose  angles  are  55°  and  80°,  construct  a  simi- 
lar triangle  one-fourth  as  large. 


CHAPTER   VIII.  -  REVIEW.  197 

4.  Given  a  right  triangle,   an  acute  angle  of  which  =  35°;   construct  a 
similar  triangle  four  times  as  large. 

5.  Construct  an  isosceles  triangle  with  the  angle  at  the  vertex  =  30°,  and 
one  side  =  6ocm.     Then  construct  a  similar  triangle  one-ninth  as  large. 

6.  Construct  a  triangle  similar  to  a  triangle  whose  sides  are  to  each  other  as 
the  numbers  7,  8,  II. 

7.  If  a  triangle  is  to  be  made  three  times  as  large  as  a  given  triangle,  but 
similar  to  it,  what  values  must  its  angles  have  ?  its  sides? 

8.  Prove  that  parallels  divide  all  lines  that  intersect  them  into  proportional 
parts. 

Hints.  —  This  is  an  extension  of  §  112.      Compare  also  this  theorem  with  the 
theorem  of  §  165. 

9.  The  sides  of  a  right  triangle  are  21™,  28™,  and  35™;    find  (z.)  the  seg- 
ments of  the  hypotenuse  made  by  a  perpendicular  let  fall  from  the 
vertex  of  the  right  angle,  (z'z.)  the  length  of  this  perpendicular. 

10.    Prove  that  the  bisector  of  an  angle  of  a  triangle  divides  the  opposite  side 
into  parts  that  have  the  same  ratio  as  the  adjacent  sides. 


Hints.  —  If  ABC  is  the  triangle,  B  D  the  bisector,  prolong  AB  till  it  is  met 

by  a  parallel  to  the  bisector  throuh  C.    A  A  B  D 
A  CBE\s,  isosceles. 


at  E  by  a  parallel  to  the  bisector  through  C.    A  A  B  D  ~  A  ACE,  and 


11.  The  three  medians  of  a  triangle  meet  in  one  point.     (A  median  is  a  line 
drawn  from  a  vertex  and  bisecting  the  opposite  side.) 

&ints.  —  LetA£Cbe  the  triangle.  Draw  two  of  the  medians  AD  and  BE 
meeting  at  a  point  O,  and  then  draw  DF\\  AC  meeting  BE  in  F.  Show 
that  A  D  F  O  ^  A  A  O  E  ;  whence  show  that  A  O  :  O  D  =2:1.  Draw 
the  third  median  CG,  cutting  AD  in  P,  and  show  in  like  manner  that 
AP  :  PD  II  2  :  i  ;  therefore  P  must  coincide  with  O. 

The  point  O  is  called  the  centre  of  gravity  of  the  triangle. 

12.  Construct  a  right  triangle  having  given,  — 

(«.)    The  hypotenuse  70°™,    and  the  ratio  3  :  2  of  the  legs. 
(z'z.)  The  altitude  40°™,  and  the  ratio  3  :  5  of  the  legs. 

13.  Construct  on  the  diagonal  of  a  rectangle  an  equal  rectangle. 

14.  What  is  the  ratio  of  the  area  ojf  a  field  to  that  of  its  plan  on  paper,  — 

(z.)    if  the  scale  of  reduction  is  icm  to  the  metei  ? 
(zz.)  if  the  scale  of  reduction  is  imm  to  the  meter  ? 

15.  Wishing  to  find  the   distance  from  a  point  A  on  the  mainland  to  an 
island  B,  I  erect  at  A  a  line  A  C  JL  A  B,  measure  upon  it  a  length  A  C  = 
900™,  and  also  measure  the  angle  ACB,  which  I  find  to  be  80°.     From 
these  data  find  the  distance  A  B. 

16.  In  Fig.  s6j,  let  A  X—  542™,  A  Y=  735™,  and  the  angle  A  =  57°  30'; 
find  the  distance  XY. 


198 


GEOMETRY    FOR   BEGINNERS. 


17.   A  vertical  rod  5™  long  throws  upon  a  horizontal  plane  a  shadow  5.7™ 

long.  At  the  same  time  a  tower 
throws  a  shadow  130™  long.  How 
high  is  the  tower  ? 

18.  Explain  how  to  find  the  height  A  B 
(Fig.  174)  by  means  of  the  right 
triangles  4£CandA£D. 

19.  Explain    how,    without    measuring 
any  angles,  the  distance  X  Y  (Fig. 
*75)  can   be    found   by  means  of 
observations     and     measurements 
performed    on    this    side    of    the 
water 

20.   The  sides  of  a  pentagon  are  12™,  20™,  nm,  15™,  and  22m  ;  the  perimeter 
of  a  similar  pentagon  is  i6m  ;    find  its  sides. 


Fig.  ///. 

21.  Construct  a  polygon  similar  to  a  given  polygon,  and  having  to  it  the 
ratio  4 :  9. 

22.  Construct  a  rectangle  similar  to  one  given  rectangle  and  equal  to  another 
given  rectangle. 


§   l8o.]  CHAPTER  IX.  —  THE   CIRCLE.  199 


CHAPTER   IX. 
THE     CIRCLE. 

CONTENTS.— I.  Sectors,  Angles  at  the  Centre,  Chords,  Segments  (§§  180-184). 
II.  Inscribed  Angles  (§§185-188).  III.  Secants  and  Tangents  (§$  189-192). 
IV.  Two  Circles  (§§  193,  194).  V.  Inscribed  and  Circumscribed  Figures 
(§§  195-207).  VI.  Length  of  a  Circumference  (§§  208-211).  VII.  Area  of 
a  Circle  (§§  212-218). 

I.  — Sectors,  Angles  at  the  Centre,  Chords,  Segments. 

§  180.   Review  §§  26,  27,  28. 

Some  new  definitions  will  now  be  given  :  — 
Definitions.  —  I.    The  parts  into  which  a  circle  (Fig.  176) 
is  divided  by  two  radii  are  called  SECTORS. 

II.  The  angle  between  two  radii  is  called  an  ANGLE  AT  THE 
CENTRE. 

III.  A  straight  line  joining  two  points  of  a  circumference  is 
called  a  CHORD. 

IV.  A  chord  passing  through  the  centre  is  called  a  DIAMETER. 

V.  A  chord  divides  the  circumference  into  two  arcs  ;  if  equal, 
these  arcs  are  called  SEMI-CIRCUMFERENCES  ;  if  unequal,  the  GREATER 
<ind  the  LESS  arcs,  respectively. 

A  chord  is  often  said  to  subtend  the  arcs  into  which  it  divides 
the  circumference. 

VI.  A  chord  divides  the  circle  into  two  parts,  called  SEGMENTS  ; 
if  equal,  they  are  called  SEMICIRCLES  ;  if  unequal,  the  GREATER  and 
LESS  segments,  respectively. 

In  Fig.  ij6  point  out  or  name  sectors,  angles  at  the  centre, 
chords,  a  diameter,  greater  and  less  arcs,  greater  and  less  seg- 
ments, semi-circumferences,  and  semicircles. 


200  GEOMETRY  FOR  BEGINNERS.  [§   180. 

Two  radii  form  at  the  centre  two  angles  (§  47,  Note  2);  but, 
unless  otherwise  stated,  the  concave  angle 
(less  than  180°)  is  always  to  be  under- 
stood. To  every  such  angle  there  corre- 
sponds a  definite  arc,  a  definite  chord,  a 
definite  sector,  and  a  definite  segment. 
Point  out  those  which  correspond  to  the 
angle  AOB  (Fig.  176). 

From  what  precedes  several  properties 
Fig.  176.  °f  tne  circle  follow  as  corollaries,  or  by 

the  method  of  superposition. 

Corollaries.  —  I.    A  circle  has  only  one  centre. 

2.  All  radii  of  a  circle  are  equal. 

3.  A  diameter  is  twice  a  radius  of  the  same  circle. 

4.  All  diameters  of  a  circle  are  equal  (why?) 

5.  An  arc  can  be  placed  on  an  equal  arc  of  the  same  circle  (or 
an  equal  circle)  so  as  to  coincide  with  it. 

6.  Circles  are  equal  whose  radii  or  whose  diameters  are  equal. 

Exercises.  —  1.  Draw  a  figure  of  your  own,  and  illustrate  the  definitions 
of  this  section. 

2.   Of  what  is  a  circumference  the  locus  ? 

3«    What  determines  (z.)  the  position,  (zz.)  the  magnitude,  of  a  circle  ? 

4.    What  three  different  positions  can  a  point  have  as  regards  the  circle  ? 

5*  Given  a  circle,  and  the  distance  of  a  point  from  the  centre ;  how  can 
you  tell  whether  the  point  will  be  inside,  in,  or  outside  the  circumference  ? 

6.  Draw  a  circle,  having  given  its  diameter. 

7.  In  a  given  circle  make  a  chord  equal  to  a  given  line. 

8.  Make  a  few  sectors  differing  in  magnitude. 

9.  Make  two  sectors  differing  in  magnitude,  but  having  the  same  angle  at 
the  centre. 

10.  What  arc  (greater  or  less)  corresponds  to  a  concave  angle  at  the 
centre  ?  to  a  straight  angle  ?  to  a  convex  angle  ?     What  segment  in  each  of 
these  cases  ? 

11.  Make  a  few  segments  (greater  and  less)  and  name  the  corresponding 
arcs  and  angles  at  the  centre. 


§  l8l.]  CHAPTER  IX. — THE   CIRCLE.  201 

12.  If  the  angle  at  the  centre  is  200°,  what  is  the  corresponding  arc  ?  the 
corresponding  segment  ? 

13.  If  the  angle  at  the  centre  is   180°,  what  is  the  corresponding  arc? 
sector  ?    segment  ? 

14.  Divide  (free-hand)  a  sector  into  two  equal  parts.     Are  the  parts  also 
sectors  ? 

15.  Divide  (free-hand)  a  segment  into  two  equal  parts.    Are  the  parts  also 
segments  ? 

16.  If  you  subtract  a  segment  from  the  corresponding  sector,  what  figure  is 
left? 

17.  Prove  that  a  diameter  is  the  longest  chord  in  a  circle.     (Use  §  62.) 

§  181.  Theorem.  —  In  the  same  circle,  or  equal  circles,  fa 
equal  angles  at  the  centre  correspond  equal  arcs,  chords,  sectors, 
and  segments. 

Draw  a  suitable  figure,  state  hypothesis  and  conclusions  (as 
concisely  as  possible  in  terms  of  the  letters  marked  on  the  figure), 
and  prove  by  superposition. 

Remark. — There  are  altogether  four  converse  theorems.  State 
them,  and  prove  them  (by  superposition) . 

Hence  it  appears  that  the  equality  of  any  one  pair  of  the  five 
elements  mentioned  in  the  theorem  implies  the  equality  of  all  the 
other  pairs. 

Corollaries.  —  I.  If  the  angle  at  the  centre  is  a  straight  angle 
(180°),  the  corresponding  arc  is  a  semi-circumference,  and  the 
corresponding  sector  is  a  segment  equal  to  a  semicircle. 

2.  Since  the  sides  of  a  straight  angle  form  a  diameter,  every 
diameter  bisects  a  circle  and  also  its  circumference. 

3.  If  the  angle  at  the  centre  is  a  right  angle,  the  corresponding 
arc  is  half  of  a  semi-circumference,  and  the  corresponding  sector 
is  half  of  a  semicircle. 

4.  Two  diameters  at  right  angles  to  each  other  divide  a  circle 
and  also  its  circumference  into  four  equal  parts. 

Definition.  —  One-fourth  of  a  circle  is  called  a  QUAPRANT. 
The  corresponding  arc  is  the  arc  of  a  quadrant. 


202  GEOMETRY    FOR    BEGINNERS.  [§   182. 

Exercises.  —  1.  If  the  angle  at  the  centre  is  1 80°,  what  is  the  arc  ?  the 
chord  ?  the  sector  ?  the  segment  ? 

2,    How  many  degrees  are  there  in  an  arc  of  a  quadrant  ? 

8*    What  is  the  ratio  of  a  quadrant  and  a  half  to  the  whole  circle  ? 

4.  In  a  given  circle  make  (*'.)  an  arc  equal  to  a  given  arc,  («.)  a  sector 
equal  to  a  given  sector,  (in.)  a  segment  equal  to  a  given  segment. 

5*    Make  an  arc  of  225°. 

6.  Make  an  arc  equal  to  twice  a  given  arc. 

7.  Make  an  arc  equal  to  five  times  a  given  arc. 

8.  Make  a  sector  equal  to  twice  a  given  sector. 

9.  Can  you  make  a  segment  equal  to  twice  a  given  segment  ? 

§  182.  If  (Fig.  177)  the  angles  AOM,  MON,  NOP,  FOB, 
COQ,  QOR,  ROD,  are  equal,  then  the 
corresponding  arcs  and  sectors  are  also 
equal  (§  181).  And  it  is  obvious  (by 
simple  comparison,  as  in  §  37)  that  the 
following  ratios  hold  true  :  — 

Angles  AOB  :  COD  =  4:3; 

Arcs         AB-.CD     =4:3; 

Sectors  A03:  COD  =  4  :$. 

This  reasoning  may  be  applied  to  any 
angles  which  have  a  common  measure,  and,  in  higher  works  on 
Geometry,  is  extended  to  incommensurable  arcs  with  the  same 
result.  Hence,  — 

Theorem.  —  In  the  same  circle,  or  equal  circles,  two  arcs,  or 
two  sectors,  have  the  same  ratio  as  the  corresponding  angles  at  the 
centre. 

Corollary. — A  sector  has  the  same  ratio  to  the  entire  circle 
that  the  corresponding  arc  has  to  the  entire  circumference. 

Exercises. — 1.    Make  two  arcs  which  shall  be  to  each  other  in  the  ratio 

2:5- 

2.   What  ratios  have  the  following  sectors  to  the  entire  circle  ? 
(«'.)    A  sector  of  30°.  (m.)  A  sector  of  90°. 

(?V.)  A  sector  of  75°.  (iv.)  A  sector  of  325°. 


§  i83.] 


CHAPTER   IX. — THE   CIRCLE. 


203 


Fig.  178. 


§  183.  Theorem  I.  —  The  diameter  perpendicular  to  a  chord 
bisects  this  chord  and  also  the  arcs  subtended  by  it. 

Proof.  —  Draw  radii  to  the  ends  of  the  chord,  and  apply  §  74. 
To  prove  that  the  arcs  are  bisected,  make  use  of  §  181  and 
Axiom  III. 

What  two  other  properties  does  this 
diameter  possess? 

Theorem  II. — A  perpendicular  erected 
at  the  middle  of  a  chord  passes  through  the 
centre  of  the  circle  (or  is  a  diameter). 

Proof.  —  This  perpendicular  passes 
through  every  point  which  is  equidistant 
from  the  ends  of  the  chord  (§  87),  and 
the  centre  of  the  circle  is  such  a  point. 

What  other  properties  does  this  perpendicular  have  ? 

Remark.  —  We  see  (Fig.  178)  that  the  middle  of  the  chord, 
the  points  of  bisection  of  the  major  and  minor  arcs,  and  the  centre 
of  the  circle  are  in  one  straight  line ;  further,  that  this  line  is  per- 
pendicular to  the  chord ;  and,  finally,  that  it  bisects  the  two  angles 
at  the  centre,  their  corresponding  sectors  and  segments.  This 
makes,  in  all,  eleven  conditions  which  the  line  CD  fulfils  (or 
eleven  properties  which  it  possesses).  They  are  so  related  that, 
if  any  two  of  them  are  known  to  be  fulfilled,  it  follows  that  all  the 
others  must  also  be  fulfilled. 

Problem.  —  To  find  the  centre  of  a  given  circle  (or  arc). 

Analysis.  —  Use  the  method  of  loci  (§  92)  in  connection  with 
Theorem  II.  of  the  present  section. 

Corollaries.  —  i.  Through  three  points  not  in  a  straight  line 
a  circle  can  always  be  drawn.  (Compare  Exercise  7,  page  102.) 

2.  Two  circumferences  that  have  three  points  in  common  coin- 
cide wholly. 

Hence,  a  circle  may  be  named  by  three  letters  at  points  on  its 
circumference. 


204  GEOMETRY  FOR   BEGINNERS.  [§   184. 

Exercises.  —  1.   Bisect  a  given  arc. 

2.  Bisect  a  given  sector. 

3.  Bisect  a  given  segment. 

4.  Find  the  centre  of  a  given  circle. 

5.  Find  the  centre  of  a  given  circle,  using  only  one  chord  and  one  per- 
pendicular. 

6.  Find  the  centre  of  a  circle  which  shall  pass  through  three  given  points. 

7 .  Given  a  curve,  how  can  you  ascertain  whether  it  is  the  arc  of  a  circle 
or  not  ? 

8.  Find  the  locus  of  the  centres  of  all  the  circles  which  pass  through  two 
given  points. 

9.  Draw  a  circle  which  shall  pass  through  two  given  points,  and  whose 
centre  shall  be  in  a  given  line.     (In  what  case  is  there  no  solution  ?     In  what 
case  is  the  problem  indeterminate  ?) 

10.  Draw  a  circle  with  a  given  radius  and  passing  through  two  given  points. 
(Two  solutions.     In  what  case  only  one  ?     In  what  case  no  solution  ?) 

11.  Find  the  locus  of  the  middle  points  of  a  system  of  parallel  chords. 

12.  Through  a  given  point  within  a  circle  draw  a  chord  which  is  bisected 
at  this  point. 

§  184.  Let  the  radius  OA  (Fig.  179)  remain  fixed,  and  con- 
ceive that  a  second  radius  turns  about  O 
as  a  centre,  starting  from  the  position 
OA,  and  coming  successively  into  the  po- 
sitions OB,  OC,  OD,  etc.  The  greater 
the  angle  at  the  centre,  AOB,AOC,AOD, 
etc.,  the  greater  the  corresponding  chord 
AB,  AC,  AD,  etc.  Likewise  the  greater 
the  chord  the  nearer  it  approaches  the 
p.  centre  O  j  and  finally,  when  the  moving 

radius   arrives   at   the   position    OE,   the 

chord  passes  through  the  centre,  and  becomes  a  diameter.     In 

this  way  we  may  arrive  at  the  following  conclusions  :  - 

Theorems.  —  I.   The  greater  the  angle  at  the  centre,  the  greater 

the    corresponding    chord   (the    angle    being  supposed  less   than 

180°). 


§   185.]  CHAPTER   IX. THE    CIRCLE.  205 

NOTE.  —  The  chord  does  not,  however,  increase  in  the  same  ratio  as  the  angle 
at  the  centre.  The  chord,  for  instance,  of  twice  an  angle  is  greater,  but  not  twice 
as  great,  as  the  chord  of  the  angle  itself. 

II.  The  diameter  is  tJie  longest  chord  of  a  circle. 

III.  Equal  chords  are  equally  distant  from  the  centre. 

IV.  Of  two  unequal  chords,  the  greater  is  nearer  to  the  centre, 
and  conversely. 

Exercise.  —  Draw  through  a  given  point  within  a  circle  the  longest  chord ; 
the  shortest  chord. 

II.  —  Inscribed  Angles. 

§  185.  Definition. — An  angle  B AC  (Fig.  r8d)  whose  vertex 
is  in  the  circumference  of  a  circle,  and  whose  sides  are  chords,  is 
called  an  INSCRIBED  ANGLE. 

To  every  inscribed  angle  corresponds  a  definite  arc  and  a  defi- 
nite chord.     The  inscribed  angle  is  often 
said  to  "  stand  upon  "  its  arc. 

By  an  angle  incribed  in  a  segment 
(DEFM)  is  understood  an  inscribed 
angle  (DEF)  whose  sides  pass  through 
the  ends  of  the  arc  of  the  segment. 

Exercises.  —  1.  Draw  a  few  inscribed  angles, 
and  point  out  the  corresponding  arcs  and  chords. 

2.  Inscribe  several  angles  in  the  same  seg- 
ment. FiS-  *8°- 

§  186.  Theorem.  — An  inscribed  angle  is  equal  to  one-half  of 
the  angle  at  the  centre  which  has  the  same  arc. 

Proof.  —  The  centre  of  the  circle  may  lie  (/.)  in  one  of  the  sides 
of  the  angle  (Fig.  181),  or  (it.)  between  the  sides  of  the  angle 
(Fig.  182),  or  (///.)  without  the  sides  of  the  angle  (Fig.  183).  So 
there  are  three  cases,  but  the  second  and  third  are  closely  con- 
nected with  the  first. 

Case  (i.).  —  Draw  the  radius  OC,  and  use  §§  66  and  80. 


206 


GEOMETRY    FOR    BEGINNERS. 


[§    186. 


Case  (ii.)-  —  Draw  the  diameter  AD,  apply  the  proof  for  Case 
/'.)  twice,  and  add  the  results. 


Fig.  181. 


Fig.  182. 


Fig.  183. 


Case  (m.)-  — Draw  the  diameter  AD,  apply  the  proof  for  Case 
(/'.)  twice,  and  subtract  the  results. 

Corollaries.  —  I.  Inscribed  angles  which  stand  upon  the  same 
arc  are  equal. 

Proof.  —  They  are  the  halves  of  the  same  angle  at  the  centre. 

2.  All  angles  inscribed  in  the  same  segment  are  equal. 

Proof.  —  They  are  inscribed  angles  standing  upon  the  same  arc  ; 
namely,  the  arc  found  by  subtracting  the  arc  of  the  segment  from 
tne  whole  circumference. 

Upon  what  arc  does  the  angle  DEF  (Fig.  180),  inscribed  in 
the  segment D  EFM,  stand? 

3.  An  angle  inscribed  in  a  semicircle  is  a  right  angle. 

Proof.  —  The  angle  at  the  centre  having  the  same  arc  is  equal 
to  two  right  angles. 

4.  The  sum  of  the  angles  inscribed,  one  in  a  greater,  the  other 
in  a  less,  segment,  is  equal  to  two  right  angles.     Why  ? 

Illustrate  by  diagrams  these  four  corollaries. 

Exercises.  —  1.  What  is  the  value  of  an  inscribed  angle  standing  upon  an 
arc  of  6o°?  153°?  320°? 

2.  What  value  has  an  angle  inscribed  in  a  segment  whose  arc  is  equal  to 
90°?  135°  180°?  270°?  300°? 

3.  Is  the  angle  inscribed  in  a  greater  segment  greater  or  less  than  90°? 
What  is  its  value  inscribed  in  a  less  segment? 


§    187.]  CHAPTER   IX.  -  THE    CIRCLE.  207 

4.  The  arcs  included  by  two  parallel  chords  are  equal.     (Two  methods  of 
proof.) 

5.  Parallel  chords  drawn  from  the  ends  of  a  diameter  are  equal.     (Use 
Corollary  3  of  this  section,  and  §  72.) 

G.    Find  the  locus  of  the  vertex  of  a  right  angle  whose  sides  pass  through 
two  given  points. 

Hints.  —  See  Corollary  3.   Make  the  line  which  joins  the  given  points  the  diame- 
ter of  a  circle. 

7.  Construct  a  right  triangle,  having  given  its  hypotenuse.     Is  the  problem 
determinate  or  not  ? 

8.  Construct  a  right  triangle,  having  given  the  hypotenuse,  and  the  altitude 
upon  the  hypotenuse  as  base.     Is  this  problem  determinate  or  not  ? 

9.  Construct  a  right  triangle,  having  given  the  middle  point  of  the  hypote- 
nuse, the  vertex  of  the  right  angle,  and  the  length  of  one  leg. 

10.  Find  the  locus  of  the  vertex  of  a  given  angle  whose  sides  pass  through 
two  given  points. 

11.  Find  the  locus  of  the  middle  points  of  all  chords  drawn  through  a  given 
point  in  the  circumference  of  a  given  circle.     (Use  §  183,  Theorem  II.,  and 
Corollary  3  of  the  present  section.) 

§  187.    If  at  any  point  D  (Fig.  184)  of  the  diameter  AB  of  a 
circle  we  erect  a  perpendicular  DC, 
and   join   AC   and   BC,    the    angle 
ACB=9o°(§    186,   Corollary   3); 
whence  it  follows,  from  §  166,  that, 
I. 


II  {C^tf  X  AD; 
'\BC2=AB  x  BD. 

Exercises.  —  1.    State  the  above  results  in  general  terms  as  two  theorems. 

2.  Verify  the  theorems  for  the  case  where  D  is  at  the  centre  of  the  circle. 
Draw  a  diagram. 

3.  If  (Fig.  184)   0£=24cm,  and  OD=  iocm,  find  DC,  AC,  and  BC. 

§  188.  Through  a  point  P  (Fig.  185)  within  a  circle  draw  any 
two  chords  APB  and  CPD.  Join  AD  and  C  B.  Show  that 
&APD  ~  A  CPB  (§§  54,  186,  161);  whence  PA:PD  = 


208 


GEOMETRY  FOR  BEGINNERS. 


[§ 


Fig.  185. 


PC:PB\  or,  multiplying  together  the  extremes  and  the  means, 

PA  x  PB  =  PC  x  PD. 
Theorem.  —  If  chords  are  drawn 
through  any  point  within  a  circle,  the 
product  of  the  two  parts  into  which  the 
chord  is  divided  by  the  point  is  the  same 
for  all  the  chords. 

Corollary.  —  Among  the  chords  which 
may  be  drawn  through  P  is  the  diameter 
EPF;  therefore,  the  constant  value  of 
the  above-mentioned  product,  for  a  given 
position  of  P,  is  the  product  of  the  two  parts  of  the  diameter  drawn 
through  P. 

Exercises.  —  1.    Prove  §  187,  Theorem  I.,  using  this  Corollary  and  §  183. 

2.  If  the  radius  of  a  circle  =  24°™,  find  the  product  of  the  parts  of  any 
chord  drawn  through  a  point  iocm  from  the  centre. 

3.  Prove  that  the  angle  APC  (Fig.  /#5),  between  two  intersecting  chords 
AB  and  CD,  is  measured  by  half  the  arc  A  EC  included  between  its  sides  + 
half  the  arc  DFB  included  between  its  sides  produced.     (Use  §  66  and 
§  186.) 

III. —  Secants  and  Tangents. 

§  189.   Definition  I. — A   straight  line   of  indefinite  length 
A  B  (Fig.  186} ,  which  cuts  the  circumference  of  a  circle  in  two 
points  is  called  a  SECANT. 

If  we  conceive  the  secant  AB  to  turn 
about  the  point  A,  towards  the  right,  the 
other    point    of  intersection  B  will  ap- 
A     proach  nearer  and  nearer  to  A,  and  will 
\    finally  coincide  with  A.     At  this  instant, 
the  secant  takes  the  position   CD,  and 
has  only  one  point  A   in  common  with 
the  circumference. 


§    IQO-]  CHAPTER   IX.— THE   CIRCLE.  209 

Definition  II.  —  A  line  which  touches  a  circumference  in  one 
point  without  cutting  it  is  called  a  TANGENT. 

Such  a  line  is  usually  said  to  touch  the  circle ;  and  the  point 
where  it  touches  the  circle  is  called  the  point  of  contact. 

Exercises.  —  1.    In  what  relation  does  a  secant  stand  to  a  chord  ? 

2.  Can  a  straight  line  cut  a  circle  in  more  than  two  points  ? 

3.  What  three  different  positions  may  a  straight  line  have  with  respect  to 
a  circle  ? 

§  190.  Letj5C  (Fig.  187)  touch  the  circle  at  the  point  A.  Join 
A  to  the  centre  O,  and  also  join  any  other 
point  of  BC,  as  Z>,  to  O.  Then  O  D> 
OA  (why?).  That  is,  OA  is  shorter 
than  any  other  line  which  can  be  drawn 
from  O  to  the  line  BC;  therefore  (§  83), 
OA  i_  BC  and  BC _L  OA. 

Theorem.  —  A  tangent  to  a  circle  is 
perpendicular  to  the  radius  drawn  to  the 
point  of  contact.  Fig.  187. 

Exercises.  —  1 .  Find  the  locus  of  the  centres  of  circles  which  touch  a  given 
line  in  a  given  point. 

2.  Find  the  locus  of  the  centres  of  circles  which  have  a  given  radius  and 
touch  a  given  line.     (Two  lines.     See  §  90.) 

3.  The  radius  perpendicular  to  a  tangent  bisects  every  chord  parallel  to 
the  tangent. 

§  191.  Problem.  —  Through  a  given  point  to  draw  a  tangent 
to  a  given  circle. 

There  are  three  cases ;  for  the  given  point  may  lie,  (*.)  within, 
(ii.)  in,  (Hi.)  without,  the  circumference. 

Case  (£.)  cannot  be  solved.    Why  ? 

Case  (ii.}. —  §  190  supplies  the  means  of  solution.  How? 
Give  the  construction. 


210  GEOMETRY   FOR   BEGINNERS.  [§   191. 

Case  (iii.). —  Construction.  —  Let  A  (Fig.  /<?£)  be  the  given 

point,  O  the  centre  of  the  given 
circle.  Join  A  O,  and  upon  A  O 
as  a  diameter  describe  a  circle 
cutting  the  given  circle  in  B  and 
C.  AB  and  A  C  are  both  tan- 
gents to  the  given  circle. 

Proof.  —  §  186,  Corollary  3, 
and  §  190. 

Corollary.  —  The    tangents 
A  B  and  A  C  (Fig.  188)  are  equal. 

Proof.  —  Show  that  the  triangles  ABO  and  A  C  O  are  equal. 

Exercises.  —  1.    Draw  a  tangent  through  a  given  point  in  the  circumfer- 
ence of  a  circle. 

2.  Draw  tangents  through  a  given  point  without  the  circumference  of  a 
circle. 

3.  If  tangents  are  drawn  from  a  point  without  a  circle,  the  line  which 
joins  the  point  to  the  centre  of  the  circle  bisects  the  angle  made  by  the  tan- 
gents. 

4.  If  the  angle  between  two  tangents  drawn  through  a  point  without  a 
circle  is  a  right  angle,  find  the  angle  at  the  centre  formed  by  radii  drawn  to 
the  points  of  contact. 

5*    Draw  a  circle  with  a  given  radius  that  shall  touch  a  given  line  in  a  given 
point.     (Two  solutions.) 

6.  Draw  a  circle  with  a  given  radius  that  shall  touch  a  given  line  and  pass 
through  a  given  point  (two,  one,  or  no  solution,  according  as  the  distance  from 
the  given  point  to  the  given  line  is  <,  =,  or  >  the  diameter  of  the  circle). 

7.  Describe  a  circle  that  passes  through  a  given  point  and  touches  a  given 
line  in  a  given  point.     (Where  cannot  the  first  point  be  ?) 

8«   To  a  given  circle  draw  a  tangent  which,  — 

(a)  is  parallel  to  a  given  line. 

(£)  is  perpendicular  to  a  given  line. 

(c)  makes  a  given  angle  with  a  given  line. 
9.   Describe  a  circle  that  touches  two  given  intersecting  lines,  and,  — 

(a)  whose  centre  is  in  a  given  line. 

(£)  which  touches  one  of  the  lines  in  a  given  point. 

(f)  which  has  a  given  radius. 


§ 


CHAPTER   IX. — THE   CIRCLE. 


211 


10.  Describe  a  circle  that  touches  two  given  parallel  lines,  one  of  them  in 
a  given  point. 

11.  Describe  a  circle  that  touches  two  given  parallel  lines,  and  passes 
through  a  point  between  them.     (Two  solutions.) 

12.  Describe  a  circle  that  touches  three  given  lines.     (Four,  two,  or  no 
solution.) 


A 
Fig.  189. 


§  192.  Let  (Fig.  189)  AD  be  a  tangent  to  a  circle  at  A,  and 
AB  any  chord  drawn  from  A.  Draw 
the  diameter  AC,  and  join  B  to  the 
centre  O  of  the  circle.  CAD  =  90° 
(why?)  =  £  arc  ABC  (measured  in 
degrees) .  CAB=±COB  (why ?) 
=  £  arc  C B  (measured  in  degrees). 
The  angle  between  the  tangent  and 
the  chord  BAD  =  CAD  —  CAB. 
Therefore,  BAD  =  £  arc  ABC  -  % 
arc  C  B  =  £  arc  BD. 

Theorem.  —  The  angle  between  a  tangent  and  a  chord  drawn 
through  the  point  of  contact,  is  measured  in  degrees  by  half  the  arc 
included  between  its  sides. 

Exercises.  — 1.  Prove  that  the  obtuse  angle  BAR  (Fig.  189)  made  by 
the  tangent  and  the  chord  is  measured  by  half  the  arc  included  between  its 
sides. 

2.  The  angle  made  by  two  secants  that  meet  without  the  circle  is  measured 
by  half  the  difference  of  the  arcs  included  between  its  sides. 

Hint. — Join  two  alternate  points  in  which  the  secants  meet  the  circle.  Use 
§  66  and  §  130. 

3.  The  angle  made  by  two  tangents  drawn  through  a  point  without  a  circle 
is  measured  by  half  the  difference  of  the  arcs  included  between  its  sides  (§  66 
and  the  Theorem  above). 

4:.  The  less  arc  between  the  points  of  contact  of  two  tangents  drawn 
through  a  point  without  a  circle  =  120°.  Find  the  angle  between  the  tan- 
gents. 

5.  The  angle  made  by  a  tangent  and  a  secant  that  meet  without  the  circle 
is  measured  by  half  the  difference  of  the  arcs  included  between  its  sides. 


212 


GEOMETRY    FOR    BEGINNERS. 


[§    193. 


IV.  ~  Two  Circles. 

§  193.  The  relative  position  of  two  circles  depends  upon  the 
positions  of  their  centres  and  the  lengths 
of  their  radii. 

Definitions.  —  I.  Two  circles  that 
have  the  same  centre  (Fig.  190)  are  said 
to  be  CONCENTRIC. 

II.  The  plane  figure  lying  between  their 
circumferences  is  called  a  RING. 

III.  Two  circles  that  have  different  cen- 
tres are  said  to  be  ECCENTRIC. 

IV.    The  line  joining  their  centres  is  called  the  LINE  OF  CENTRES. 

§  194.  Let  (Fig.  191}  O  and  P  be  the  centres  of  two  eccentric 
circles,  with  the  radii  r  and  r1  respectively,  and  let  us  proceed  to 
examine  the  different  positions  which  these  circles  will  have,  as  the 
distance  of  their  centres  OP  =  c  is  supposed  continually  to  in- 
crease. There  are  five  different  cases. 

Case  (i.)  (Fig.  192} .  c  <  r  —  r'.  If  the  distance  of  the  centres 
is  less  than  the  difference  of  the  radii,  the  smaller  circle  lies  wholly 
within  the  larger. 


Fig.  191. 


Fig.  192. 


193- 


Case  («*.)  (Fig.  193) .  c=r—  r1.  If  the  distance  of  the  cen- 
tres is  equal  to  the  difference  of  the  radii,  the  circles  touch,  or  are 
tangent,  internally. 


1  94.] 


CHAPTER   IX.  —  THE    CIRCLE. 


213 


Case  (Hi.'}  (Fig.  193)  .  c  >  r  —  r1  and  <  r  +  /.  If  the  distance 
of  the  centres  is  greater  than  the  difference,  but  less  than  the  sum 
of  the  radii,  the  circles  intersect  in  two  points. 

Case  (if.)  (Fig.  194)  .  c  =  r  +  /.  If  the  distance  of  the  cen- 
tres is  equal  to  the  sum  of  the  radii,  the  circles  touch,  ^  or  are 
tangent,  externally. 


Fig.  194. 


Fig.  195. 


Case  (v.)  (Fig.  195}.  c>r+r'.  If  the  distance  of  the  cen- 
tres is  greater  than  the  sum  of  the  radii,  the  circles  lie  wholly 
outside,  each  other. 

Exercises.  —  1.  Can  you  give  a  reason  why  two  circles  cannot  meet  in 
more  than  two  points  ? 

2.  What  different  positions  can  two  circles  have  relatively  to  each  other  ? 

3.  Let  r  and  r1  be  the  radii  of  two  circles,  c  the  distance  of  their  centres. 
What  relative  positions  have  the  circles  for  the  following  values  of  r,  r'  ,  and  c  ? 

(«)r=s,  ^=3,  c  =  8. 
0)  r—T,  r'-^,  c  =  2. 
(c}  r  =  6,  rr=2,  <r=io. 
(</)r=8,  r'=3,  <:=  6. 
0)  ^=9,  r'=5,  r=  4. 

4.  Describe,  with  the  radii  35cm  and  25c 
each  other  (z.)  internally,  (zV.)  externally. 

5.  In  a  given  circle  make  two  circles  such  that  they  shall  both  touch  the 
given  circle  internally,'  and  also  touch  each  other  externally. 

6.  With  three  given  radii,  m,  n,  and  /,  describe  three  circles  which  shall 
mutually  touch  one  another  externally. 

Hint.  —  First  construct  a  triangle  having  the  sides  m  +  n,  m  +  p,  and  n  -\-  p. 


(/) 

r  = 

6, 

r/_ 

6, 

^=   8. 

(d 

?*  ^^ 

10, 

r'= 

3, 

ir=    7. 

(A) 

r  = 

5» 

r/_ 

5» 

f  =  10. 

(0 

r  — 

12, 

r'= 

7. 

r=    9. 

(/§) 

r  = 

9, 

r/_ 

2, 

c=  5- 

two  circles  which  shall  touch 


214 


GEOMETRY   FOR   BEGINNERS. 


[§ 


V.— Inscribed  and  Circumscribed  Figures. 

§  195.  Definitions.  —  I.  A  circle  is  said  to  ^INSCRIBED  in  a 
polygon  when  its  circumference  touches  all  the  sides  of  the  polygon  ; 
and  the' polygon  is  said  to  be  CIRCUMSCRIBED  about  the  circle  (Fig. 
196). 

II.  A  circle  is  said  to  be  CIRCUMSCRIBED  about  a  polygon  when 
its  circumference  passes  through  all  the  corners  of  the  polygon  ; 
and  the  polygon  is  said  to  be  INSCRIBED  in  the  circle  (Fig.  197) . 

In  Fig.  196,  the  sides  of 
the  polygon  are  tangents 
to  the  circle ;  in  Fig.  197, 
they  are  chords  of  the  cir- 
cle. 


Fig.  196. 


Fig.  197. 


NOTE.  —  In  treating  this  sub- 
ject we  shall  first  suppose  the  poly- 
gon to  be  given,  and  the  circle 
inscribed  or  circumscribed  ($§ 

196-201)  ;    and  then  suppose  the   circle  to  be  given  and  the  polygon  to  be 

inscribed  or  circumscribed  (§§  202-207). 


§  196.   Problem.  —  To  inscribe  a  circle  in  a  given  triangle 
ABC  (Fig.  198}. 

Construction.  —  Bisect  the  angles  A  and  B,  and  from  the  inter- 
section O  of  the  bisectors  draw 
OD1.AB.  O  is  the  centre  of 
the  required  circle,  OD  the  ra- 
dius. 

Proof.  —  Draw  O  E  J_  A  C, 
O  F±  B  C,  and  j  oin  O  C.  Show 
that  A  A  OD  ^  A  A  O  E,  and 
A  B  O  D  ^  A  B  O  F,  whence 

A  D  *     OD  =  OE  =  OF;   therefore, 

the  circle  with  O  as  centre  and 
OD  as  radius  touches  the  sides  of  the  triangle  in  D,  E,  and  F. 


§   19  7«]  CHAPTER   IX. THE    CIRCLE.  215 

Compare  this  problem  with  §  92,  Exercise  8. 

Corollary.  —  A  circle  can  always  be  inscribed  in  a  triangle. 

§  197.  Problem.  —  To  circumscribe  a  circle  about  a  given  tri- 
angle. 

Analysis,  by  means  of  §  §  87  and  92. 

Draw  a  figure,  and  give  the  construction. 

Compare  this  problem  with  §  92,  Exercise  7. 

Corollary. — A  circle  can  always  be  circumscribed  about  a 
triangle. 

Exercise.  —  Prove  that  if  a  triangle  is  equilateral,  the  radius  of  the  circum- 
scribed circle=:  twice  the  radius  of  the  inscribed  circle  (see  §  112,  Exercise). 

§  198.  Theorem. — If  a  quadrilateral  A  B  C D  (Fig.  199)  is 
circumscribed  about  a  circle,  the  sum  of  two  opposite  sides  is  equal 
to  the  sum  of  the  other  two  sides. 

Proof.  —  Apply  §  191,  Corol- 
lary ;  then  add  the  equal  parts 
of  the  sides  together,  obtaining, 
for  the  result,  AB  +  DC=  BC 
-f  AD. 

Corollary.  —  If  in  a  quadri- 
lateral the  sum  of  two  opposite 
sides  is  equal  to  the  sum  of  the 
other  two  sides,  a  circle  can  be  in-  "  F- 

scribed  in  the  quadrilateral. 

Exercises.  —  1.    How  is  the  centre  of  the  inscribed  circle  found  ? 

2.  In  what  two  kinds  of  quadrilaterals  can  a  circle  always  be  inscribed  ? 

3.  Draw  a  rhombus,  and  inscribe  in  it  a  circle. 

§  199.  Theorem.  —  If  a  quadrilateral  is  inscribed  in  a  circle, 
the  sum  of  the  opposite  angles  is  equal  to  180°. 

Proof.  —  Draw  a  figure  of  an  inscribed  quadrilateral,  then  draw 
a  diagonal,  and  apply  §  186,  Corollary  4. 


216  GEOMETRY    FOR   BEGINNERS.  [§   2OO. 

Corollary.  —  If  in  a  quadrilateral  the  sum  of  two  opposite 
angles  =  180°,  a  circle  can  be  circumscribed  about  the  quadri- 
lateral. 

Exercises.  —  1.    How  is  the  centre  of  the  circumscribed  circle  found  ? 

2.  About  what  two  kinds  of  quadrilaterals  can  a  circle  always  be  circum- 
scribed ? 

3.  Draw  a  rectangle,  and  circumscribe  about  it  a  circle. 

§  200.  Theorem. — A  circle  can  always  be  inscribed  in  a 
regular  polygon. 

Proof.  —  Use  §  119.  What  is  the  radius  of  the  circle,  and  what 
point  is  its  centre  ? 

Exercises.  —  1.    Show  how  to  find  the  centre  of  the  inscribed  circle. 

2.  Inscribe  a  circle  in  a  regular  hexagon. 

3.  Within  a  square  describe  four  circles  touching  each  other,  and  also  the 
sides  of  the  square.     What  is  the  common  radius  of  the  circles  ? 

§  201.  Theorem.  — A  circle  can  always  be  circumscribed  about 
a  regular  polygon. 

Proof.  —  Use  §  119.  What  is  the  radius  of  the  circle,  and  what 
point  is  its  centre  ? 

Exercises.  —  1.    Show  how  to  find  the  centre  of  the  circumscribed  circle. 

2.  Circumscribe  a  circle  about  a  regular  hexagon. 

3.  Circumscribe  a  circle  about  a  regular  octagon. 

§  202.  In  order  to  inscribe  regular  polygons  in  circles,  or  to 
circumscribe  them  about  circles,  it  is  necessary  to  divide  the  cir- 
cumference of  the  circle  into  as  many  equal  parts  as  there  are 
sides  in  the  polygon. 

Problem.  —  To  divide  with  ruler  and  compasses  the  circumfer- 
ence of  a  circle  into  equal  parts. 

This  problem  can  be  solved  only  in  the  following  cases  :  — 

Case  1.  —  Two,  four,  eight,  sixteen,  thirty- two,  etc.,  equal  parts. 


§    202.] 


CHAPTER   IX. THE    CIRCLE. 


217 


Solution.  —  A  diameter  bisects  the  circumference  (§  181,  Corol- 
lary 2);  two  diameters  perpendicular  to  each  other  divide  it  into 
four  equal  parts  (§  181,  Corollary  4);  repeated  bisection  of  the 
arcs  will  divide  it  into  eight,  sixteen,  thirty-two,  etc.,  equal  parts. 

Draw  a  circle,  and  divide  the  circumference  into  two  equal  parts ; 
four  equal  parts  ;  eight  equal  parts. 

Case  2.  —  Three,  six,  twelve,  twenty-four,  forty-eight,  etc.,  equal 
parts. 

Solution.  —  If  we  construct  in  a 
circle  (Fig.  200)  an  equilateral  tri- 
angle OAB,  having  each  side  equal 
to  the  radius,  it  is  clear  that,  since 
AOB  =  60°,  the  arc  AB  =  £  the 
circumference.  Therefore,  the  ra- 
dius applied  as  a  chord  six  times 
to  the  circumference  will  divide  it 
into  six  equal  parts. 

The  alternate  points  of  division, 
as  Ay  C,  E,  divide  the  circumference 
into  three  equal  parts  (Axiom  II.) . 

How  can  the  circumference  be  divided  into  twelve  equal  parts  ? 
into  twenty-four?  into  forty-eight? 

Divide  a  circumference  into  six  equal  parts  ;  twelve  equal  parts. 

Case  3.  —  Five,  ten,  twenty,  forty, 
eighty,  etc.,  equal  parts. 

Solution  (Fig.  201) .  —  Bisect  A  O 
in  C,  and  with  C  as  centre,  and  a 
radius  equal  to  CD,  describe  an  arc 
cutting  OB  in  E ;  OE  applied  as  a 
chord  ten  times  to  the  circumference 
will  divide  it  into  ten  equal  parts,  and 
DE  applied  as  a  chord  five  times 
will  divide  it  into  five  equal  parts. 


Fig.  200. 


218  GEOMETRY    FOR    BEGINNERS.  [§   203. 

The  proof  that  this  solution  is  correct  cannot  be  given  here. 

How  will  the  division  into  ten  equal  parts  enable  us,  without  the 
use  of  DjE,  to  make  the  division  into  five  equal  parts? 

How  can  a  circumference  be  divided  into  twenty  equal  parts  ? 

Divide  a  circumference  into  five,  and  into  ten  equal  parts. 

Remark.  —  With  the  aid  of  a  protractor  (see./7^.  46)  a  cir- 
cumference may  be  divided  with  sufficient  exactness  for  all  practical 
purposes  into  any  number  of  equal  parts..  Suppose,  for  instance,  that 
the  number  of  equal  parts  is  25  ;  then  the  angle  at  the  centre  cor- 
responding to  each  part  =  ^ —  =  14°  24'.  Construct  with  the  pro- 
tractor, placed  so  that  its  centre  shall  coincide  with  the  centre  of 
the  circle,  an  angle  of  14°  24' ;  the  arc  of  the  given  circle  con- 
tained between  the  sides  of  this  angle  will  be  ^V  the  entire  cir- 
cumference. 

NOTE.  —  An  angle  of  i°  (among  others)  cannot  be  constructed  with  the  ruler 
and  compasses.  The  question  may  here  occur :  how  is  a  protractor  itself  gradu- 
ated into  degrees  ?  One  way  is  as  follows :  divide  with  ruler  and  compasses  the 
entire  circumference  into  six  equal  parts,  and  then  into  five  equal  parts.  The  dif- 
ference between  two  of  these  parts  (i  —  ^)  =  -J^,  the  circumference  =  J^  the  semi- 
circumference.  Bisect  twice  in  succession  each  of  these  fifteen  parts ;  this  divides 
the  semi-circumference  into  sixty  equal  parts ;  finally,  each  of  these  sixty  parts  is 
trisected  by  repeated  trials,  opening  or  closing  a  little  the  compasses  till  the  right 
opening  is  obtained.  But  the  most  perfect  method  of  graduating  a  circle  into 
degrees  and  subdivisions  of  a  degree  is  by  means  of  nicely-adjusted  screw-motion 
in  machines  made  for  this  express  purpose. 

Exercises.  —  1.  Find  all  the  angles  between  o°  and  360°  which  can  be 
constructed  by  Case  I. 

2.  Find  all  the  angles  which  can  be  constructed  by  Case  2. 

3.  Find  all  the  angles  which  can  be  constructed  by  Case  3. 

4.  What  angle  can  be  trisected  with  ruler  and  compasses  ?     Trisect  it. 

5.  Divide  with  the  protractor  a  circumference  into   seven,  nine,  fifteen 
equal  parts. 

§  203.  Problem.  —  To  inscribe  in  a  given  circle  a  regular 
polygon  {Fig.  202). 

Construction.  —  Divide  the  circumference  into  as  many  equal 


§   204.] 


CHAPTER   IX. — THE   CIRCLE. 


parts  as  the  polygon  has  sides,  and  join  in  order  the  points  of, 
division  by  straight  lines. 

Proof. —  The  sides  AB,BC,  etc., 
are  equal,  and  the  angles  ABC, 
BCD,  etc.,  are  equal  (§  181,  Re- 
mark) .  Therefore,  the  inscribed  fig- 
ure is  a  regular  polygon  (§  118). 

Remark.  —  By  means  of  the  fol- 
lowing construction,  the  side  of  any 
inscribed  regular  polygon  (except  the 
triangle  and  the  square)  can  be  found 
to  a  close  degree  of  approximation.  Fis- 202- 

Divide  the  diameter  AB  (Fig.  203)  of  the  given  circle  into  as  many 
equal  parts  as  there  are  sides  in 
the  polygon  to  be  inscribed. 
Draw  CD  ±.AB.  Prolong  A  B 
to  E  and  CD  to  F,  making  A  E 
=  DF=  one  of  the  equal  parts. 
Join  EFy  and  finally  join  the 
intersection  G  of  EF  with  the 
circumference  to  the  third  point 


of  division  H  of  the  diameter ; 

GH  is  the  side  of  the  polygon  required. 


Fig.  203. 


Exercises.  —  Inscribe  in  a  given  circle, — 

1.   A  square.  4,    A  regular  polygon  of  seven  sides. 

£•   A  regular  pentagon.  5.    A  regular  polygon  of  nine  sides. 

3.   A  regular  octagon.  6.    A  regular  polygon  of  eleven  sides. 

§  204.  Problem.  —  To  circumscribe  about  a  given  circle  a  reg- 
ular polygon  (Fig.  204). 

Construction.  —  Divide  the  circumference  into  as  many  equal 
parts  as  the  polygon  has  sides,  and  draw  tangents  through  all  the 
points  of  division. 


220 


GEOMETRY    FOR    BEGINNERS. 


[§   205 


Proof.  —  Show  that  the  quadrilat- 
erals ABCO,  CDEO,  etc.,  are  equal 
(by  the  method  of  superposition) . 

Exercises.  — Circumscribe  about  a  given 
circle, — 

1.  An  equilateral  triangle. 

2.  A  regular  hexagon. 

3.  A  regular  heptagon. 

4.  A  regular  nonagon. 

5.  A  regular  decagon. 

6.  A  regular  dodecagon. 


§  205.    Regular  polygons  are  most  easily  constructed  by  employ- 
ing circles. 

Problem.  —  To  construct  a  regular  polygon,  having  given  the 

number  of  sides,  and  the  length  AB 
{Fig.  2Oj)  of  one  side. 

Solution.  —  Find  the  centre  O  of 
the  polygon  (§  121),  and  about  O, 
with  a  radius  =  OA,  describe  a  cir- 
cle •  then  apply  AB  to  the  circum- 
ference as  many  times  as  there  are 
sides  in  the  polygon. 

Remark.  —  Every  regular  poly- 
gon is  composed  of  a  series  of  equal 
right  triangles  (§  119,  Corollary  2). 
The  sides  of  one  of  these  triangles, 

AOD  (Fig.  205),  are  the  greater  radius,  the  less  radius,  and 
the  half-side  of  the  polygon.  It  follows,  from  §  142,  that  these 
sides  cannot  have  any  values  which  we  may  choose  to  give  them, 
but  are  so  related  that  ~AO*  =  ~AI?  +  ~OL? .  The  following  table 
gives  the  values  (near  enough  for  most  purposes)  of  the  less 
radius  and  the  side,  the  greater  radius  in  each  case  being  taken 
equal  to  100. 


Fig.  205. 


§     205.] 


CHAPTER   IX. — THE   CIRCLE. 


221 


Polygon 
of 

Greater 
Radius. 

Less  Radius. 

One  Side. 

3  sides 

100 

5° 

173 

4  sides 

loo 

71 

141 

5  sides 

100 

8l 

Il8 

6  sides 

100 

87 

IOO 

7  sides 

100 

90 

87 

8  sides 

100 

92 

77 

9  sides 

IOO 

94 

68 

10  sides 

TOO 

95 

62 

12  sides 

IOO 

96 

52 

Exercises.  —  Upon  a  line  2Ocm  long  construct, — 

1.  A  regular  pentagon.  3.    A  regular  octagon. 

2.  A  regular  hexagon.  4.   A  regular  decagon. 
Construct  a  regular  polygon,  having  given,  — 

6.   Perimeter  =  i.2m,  number  of  sides  =  8. 

6.  Greater  radius  =  6ocm,  less  radius  =  3Ocm. 

7.  Greater  radius  =  5Ocm,  one  side  =  26cm. 

8.  Less  radius  =  27cm,  one  side  =  39cm. 

9.  Greater  radius  =  one  side. 

10.  One  side  =  36°™,  one  angle  =  135°. 

11.  Greater  radius  =  45cm,  angle  at  the  centre  =  30°. 

12.  Less  radius  =  32cm,  angle  at  the  centre  =  40°. 

13.  Since  a  regular  polygon  is  composed  of  a  series  of  equal  right  triangles, 
any  two  parts  which  determine  one  of  these  triangles  will  determine  the  entire 
polygon.     What  are  the  six  parts  of  one  of  these  triangles  ?     What  two  parts 
can  always  be  found  if  the  number  of  sides  is  known  ?     How  ? 

14.  Give  all  the  cases  of  two  parts  that  determine  a  regular  polygon. 

15.  The  radius  of  a  circle  =  iocm.     Find  the  side  of  the  inscribed  equilate- 
ral triangle  ;   also  a  side  of  the  inscribed  regular  decagon. 

16.  The  perimeter  of  a  square  =  4Ocm.     Find  the  radii  of  the  inscribed 
and  of  the  circumscribed  circles. 

17.  In  a  regular  hexagon  the  line  which  joins  the  middle  points  of  two 
opposite  sides  is  2m  long.     Required  the  perimeter  of  the  hexagon. 

18.  A  square,  whose  perimeter  =  40°™,  is  changed  into  a  regular  octagon 
by  cutting  isosceles  right  triangles  from  its  corners.     Find  the  sides  of  one  of 
Ihese  triangles. 


222 


GEOMETRY   FOR   BEGINNERS. 


[§  206. 


19.  One  side  of  a  marble  slab  in  the  shape  of  a  regular  octagon  =  42°™. 
Find  the  radius  of  the  inscribed  circle. 

20.  In  a  circle  whose   radius  =  I5cm,  a  regular  hexagon  and  a  regular 
dodecagon  are  inscribed.     Find  the  difference  of  their  perimeters. 

§  206.   Problem.  — Having  given  the  side  A  B  (Fig.  206}  of  an 
inscribed  regular  polygon,  to  construct  the  side  of  an  inscribed  regu- 
lar polygon  having  twice  as  many 
sides,  and  to  compute  its  length. 

Construction.  —  Bisect  the  arc 
AB  in  D,  and  join  AD.  AD 
is  the  side  of  the  polygon  re- 
quired. 

Proof.  —  Apply  §  181. 
Computation.  —  If  AB   and 
AO  are  known,  then  in  the  right 
triangle  AOC  we  can  compute 
the  value  of  CO  (how?).     Sub- 
tract CO  from  DO,  and  we  ob- 
tain CD;  lastly,  in  the  right  tri- 
angle A  CD,  in  which  AC  and  CD  are  known,  compute  AD 
(how?). 

Exercises.  —  1.    In  a  circle  whose  radius  =  im  a  regular  hexagon  is  in- 
scribed.    Find  the  side  and  the  perimeter  of  the  regular  inscribed  dodecagon. 


Solution.—],  n  this  case,  A  B  {Fig.  206}  =A  O  =  im  ;  A  C=  0.5™  ;  CO  - 

Vi2  —  o.S2  =  0.86602541";   CD  =  DO  —  CO=i  —  0.8660254  =  o.  1339746™  ;  A  D 


=  Vo.52  +  o.  1339746*  =  0.51763818^.     The  perimeter  of  the  do- 
decagon =  12  X  0.51763818™  =  6.2ii658m. 

2.  Find  the  area  of  the  triangle  A  DB  (Fig.  206}. 

3.  If  the  radius  of  the  circle  =  im,  compute  the  side  and  the  perimeter  of 
the  inscribed  regular  polygon  of  twenty-four  sides. 

The  radius  of  a  circle  =  iom.     Find,  — 

4.  The  perimeter  and  the  area  of  the  inscribed  square. 

5.  The  perimeter  and  the  area  of  the  inscribed  regular  octagon. 


CHAPTER   IX. — THE   CIRCLE. 


223 


§  207.  Problem. — Having  given  the  side  AB  (Fig.  207)  of 
an  inscribed  regular  polygon,  to  construct  the  side  of  the  circum- 
scribed regular  polygon  having 
the  same  number  of  sides,  and 
to  compute  its  length. 

Construction.  —  Bisect  the  arc 
AB  in  D ;  through  D  draw  a 
tangent,  and  prolong  it  until  it 
meets  OA  and  OB  prolonged 
in  E  and  in  F.  EF  is  the  side 
of  the  polygon  required. 

Proof  —  Conceive  the  same 
construction  carried  out  for  all 
the  equal  arcs  subtended  by  the 
sides  of  the  inscribed  polygon  ;  then  prove  that  the  triangles  OEF, 
OFG,  etc.,  are  equal  (by  showing  that  the  right  triangles  which 
are  their  halves  are  all  equal). 

Computation.  —  AB  and  EF  are  parallel,  both  being  perpen- 
dicular to  OC;  therefore,  &AOB  ^  &EOF,  and  OC ':  OD  = 
AB  :  EF.  In  this  proportion,  OD  and  A  B  are  given,  and  OC 
can  be  found  from  the  right  triangle  AOC  (how  ?) .  E  F  can  then 
be  computed  (how?). 

Exercises.  — 10  In  a  circle  whose  radius  =  im  a  regular  hexagon  is  in- 
scribed. Find  the  side  and  the  perimeter  of  the  regular  circumscribed 
hexagon. 

Solution.  —  If  (Fig.  207}  OD=i™,  then  AI?=i™,zr\d  OC= V^?*  —  AC1  — 
V*2  ~"  °-5~  =  0.8660254™  ;  hence  0.8660254 :  i  =  I :  E  F.  E  F—  1. 1547005™  =  one 
side  of  the  circumscribed  polygon.  And  the  perimeter  =  6  X  1.1547005™  =  6.928203™. 

2.  In  a  circle  whose  radius  =  im  the  side  of  the  inscribed  regular  dodeca- 
gon has  been  found  (§  206)   equal  to  0.5176318™.     Find  the  side  and  the 
perimeter  of  the  circumscribed  regular  dodecagon. 

The  radius  of  a  circle  =  iom.     Find,  — 

3.  The  perimeter  and  the  area  of  the  circumscribed  square. 

4.  The  perimeter  and  the  area  of  the  circumscribed  regular  octagon. 


224 


GEOMETRY    FOR    BEGINNERS. 


[§  208. 


VI.  — Length  of  a   Circumference. 

§  208.  In  order  to  find  the  length  of  a  circumference,  we  make 
use  of  two  general  truths,  which  are  sufficiently  obvious  from  what 
precedes. 

I.  The  perimeter  of  an  inscribed  regular  polygon  is  always  less, 
and  that  of  a  circumscribed  regular  polygon  is  always  greater,  than 
the  length  of  the  circumference. 

II.  The  greater  the  number  of  sides,  the  nearer  the  perimeter  in 
either  case  approaches  to  the  length  of  the  circumference. 

If  the  radius  of  the  circle  =  i,  the  perimeter  of  the  inscribed 
regular  hexagon  =  6  (§  202,  Case  2),  and  that  of  the  circum- 
scribed regular  hexagon  =  6.928203  (§  207) ;  therefore,  the  length 
of  the  circumference  is  more  than  6,  but  less  than  6.928203,  times 
the  length  of  the  radius. 

If  now  we  inscribe  and  circumscribe  regular  polygons  of  twelve 
sides,  twenty-four  sides,  forty-eight  sides,  etc.,  and  compute  to  the 
sixth  decimal  place  the  values  of  their  perimeters,  we  shall  obtain 
the  following  results  :  — 


Number  of 

PERIMETER  OF  THE  POLYGON. 

Sides  in 
the  Polygon. 

Inscribed. 

Circumscribed. 

6 

6.000000 

6.928203 

12 

6.211658 

6.430782 

24 

6.265257 

6.319320 

48 

6.278700 

6.292172 

96 

6.282066 

6.285430 

192 

6.282905 

6.283746 

384 

6.283115 

6.283325 

768 

6.283168 

6.283220 

1536 

6.283181 

6.283194 

3072 

6.283183 

6.283187 

§   208.]  CHAPTER   IX. THE   CIRCLE.  225 

Since  the  length  of  the  circumference  always  lies  between  the 
perimeters  of  the  inscribed  and  circumscribed  polygons  having  the 
same  number  of  sides,  and  the  perimeters  of  these  polygons  having 
3072  sides  agree  in  value  in  the  first  five  decimal  places,  therefore 
the  number  6.28318  expresses  the  length  of  the  circumference 
correctly  to  five  decimal  places ;  that  is,  the  length  of  the  circum- 
ference is  6.28318  times  the  length  of  the  radius,  or  3.14159  times 
the  length  of  the  diameter. 

The  number  which  expresses  the  ratio  of  the  circumference  of 
a  circle  to  its  diameter,  is  often  denoted,  for  the  sake  of  conve- 
nience, by  the  Greek  letter  TT  l ;  that  is,  TT  =  3.14159,  very  nearly. 

For  most  practical  purposes  we  may  take  T:  =  3^  =  ^2-,  or,  deci- 
mally, -  =  3.14.  When  greater  accuracy  is  required,  more  decimal 
places  must  be  used.  If  more  than  five  decimals  are  desired, 
they  may  be  obtained  by  continuing  the  above  process,  or  much 
more  easily  by  means  of  other  methods  known  to  mathematicians. 

But  whatever  method  be  used,  the  exact  value  of  the  ratio  of  the 
circumference  of  a  circle  to  the  diameter  cannot  be  found.  The 
farther  the  computation  is  carried,  the  nearer  we  approximate  to 
the  true  value,  but  the  series  of  decimals  which  we  obtain  never 
comes  to  an  end.  This  amounts  to  saying  that  the  circumference 
of  a  circle  and  its  diameter  are  two  incommensurable  lines  (§  38). 

NOTE.  —  The  problem  of  finding  the  ratio  between  the  circumference  of  a  circle 
and  its  diameter  exercised  the  minds  of  geometers  in  very  early  times.  In  an  ancient 

Brahmin  work,  called  Ayeen  Akbery,  the  value  is  assigned  as  2i?Z  =  3.1416.    Archi- 

1250 

medes,  the  greatest  geometer  of  antiquity  (died  212  B.C.),  using  the  method  of 
inscribed  and  circumscribed  polygons,  up  to  polygons  of  96  sides,  found  that 
the  value  lay  between  3^  and  3^.  Ludolph  of  Cologne,  with  prodigious  labor, 
carried  the  computation  to  polygons  of  32,212,254,720  sides,  which  gives  the  value 
correct  to  35  decimal  places.  Metius  found  it  to  be  355,  which  is  correct  to  six 

decimal  places.  And  Professor  Richter  of  Elbing,  Germany,  has  computed  the 
value  of  TT  to  500  decimal  places ! 

But  there  is  no  practical  advantage  in  obtaining  more  than  the  first  five  or  six 
places ;  the  value  TT  =  3.14159  is  so  near  the  truth  that  the  error  made  in  computing 

l  TT  (pronounced  like  /)  is  the  first  letter  of  a  Greek  word  meaning  circumference. 


226  GEOMETRY    FOR    BEGINNERS.  [§   2 09. 

the  circumference  of  a  circle  with  a  diameter  of  one  mile  is  considerably  less  than 
one  inch.  Accordingly,  this  problem,  which  engaged  the  attention  of  geometers 
so  deeply  when  their  methods  of  approximation  were  less  perfect,  has  now  sunk 
to  the  rank  of  those  useless  questions  over  which  only  persons  ignorant  of  geo- 
metrical science  are  willing  to  waste  their  time. 

§  209.  Let  r,  d,  and  c  denote  the  numbers  which  express,  in 
terms  of  the  same  unit,  the  lengths  of  the  radius,  the  diameter,  and 
the  circumference,  respectively,  of  a  circle ;  then,  from  what  pre- 
cedes, — 

C  =  ir(l=  2irr.  [lO.] 

Read  this  formula  in  words.  If  r  is  given,  how  can  c  be  found  ? 
If  c  is  given,  how  can  r  and  d  be  found? 

For  example,  if  r=4m,  then  d=  8m,  and  c  =  8  X  3.14  =  25.12™. 

Conversely,  if  <•=  20m,  then  d—  -  -  =  6.36™,  and  r=  3.i8m. 

3-J4 

It  is  obvious,  from  Formula  [10],  that  if  either  r  or  </be  doubled, 
trebled,  quadrupled,  etc.,  c  will  be  doubled,  trebled,  quadrupled; 
etc. ;  in  other  words,  — 

The  circumferences  of  two  circles  are  to  each  other  as  their  radii, 
or  as  their  diameters. 

§  210.  An  arc  may  be  measured  in  degrees,  minutes,  and  sec- 
onds (angular  measure),  or  its  length  may  be  expressed  (like  that 
of  any  line)  in  meters,  feet,  or  other  linear  units. 

It  follows,  from  §  181,  that,  - 

Length  of  an  arc  :  whole  circumference  =  corresponding  angle  at 
the  centre :  j6o°. 

By  means  of  this  proportion  such  questions  as  the  following  may 
be  solved  :  — 

i.    In  a  circle  whose  radius  =  5™,  find  the  length  of  an  arc 

of  45°. 

The  circumference  =  iox  3. 14  =  31. 4™.    Length  of  arc  of  45°  : 

31.4  =  45°  :  360°.    Length  of  the  arc  =  3I'445  =  3-92m. 


§  21  lo]  CHAPTER   IX.  —  THE   CIRCLE.  227 

2.  In  the  same  circle,  how  many  degrees,  etc.,  are  there  in  an 
arc  6m  long? 

The  above  proportion  in  this  case  becomes,  6  :  31.4  =  No. 
of  degrees  in  the  arc  :  360°.  No.  of  degrees  in  the  arc  = 


31-4 


=68 


§  211.   PRACTICAL  EXERCISES. 

(  When  not  otherwise  specified,  take  TT  =  37-) 

Find  the  circumference  of  a  circle,  if  the  radius  is  equal  to,  — 

1.  7m.  4.   6T3Tft.  7.   3-27m  (7r  = 

2.  236™.  5.   #".  8.   75«n(7r  =  3-14159). 

3.  4.38km.  6.   ^milc.  9.   8425*™  (^  =  3.1415926). 
Find  the  radius  and  the  diameter  of  a  circle,  if  the  circumference  is  equal 

to,— 

10.  I7m.  13.  I5ift.  16.  9.74™  (TT=  3.14). 

11.  162.4™.  14.  iiin.  17.  64cm  (TT=  3.14159). 

12.  2.16*".  15.  imile.  18.  40^0=3.1415926). 

19.  How  many  revolutions  does  a  car-wheel,  the  diameter  of  which  =  1.4™, 
make  in  going  a  distance  of  I32km? 

Ans.  Circumference  of  wheel  =  —      —  =  4.4m;   No.  of  revolutions  =  —  !  — 

7  4-4 

=  30,000. 

20.  If  the  driving-wheels  of  a  locomotive  have  a  radius  of  3ft,  how  many 
revolutions  will  they  make  in  going  from  New  York  to  Boston,  a  distance  of 
236  miles? 

21.  The  diameter  of  a  wheel  =  75cm,  and  in  going  a  certain  distance  it  was 
observed  to  make  3000  revolutions.     Required  the  distance. 

22.  What  is  the  diameter  of  a  circular  reservoir,  if  a  man  in  walking  around 
it  makes  840  paces,  and  each  pace  =  0.625™? 

23.  What  must  be  the  diameter  of  a  round  dining  table  for  eight  persons, 
if  77cm  is  allowed  to  each  person  ? 

24.  How  many  nails  are  required  to  fasten  the  cloth  covering  about  the 
edge  of  a  circular  table  1.4™  in  diameter,  the  distance  between  two  nails  to 
be  4cm? 

25.  A  toothed  wheel  3-756m  in  diameter  has  360  teeth.    Find  the  distance 
between  the  centres  of  two  teeth. 


228  GEOMETRY   FOR   BEGINNERS.  [§   211. 

26»   On  the  circumference  of  a  wheel  are  36  teeth,  and  the  distance  between 
the  centres  of  two  teeth  =  i8mm.     Find  the  diameter  of  the  wheel. 

27.  The  radii  of  three  concentric  circles  are  i,  2,  and  3m.     Find  their 
circumferences. 

28.  The  diameters  of  two  concentric  circles  are  I2cm  and  2Ocm.     Find  the 
diameter  of  a  circle  whose  circumference  lies  half-way  between  the  other 
two. 

29.  The  diameters  of  two  concentric  circumferences  are  8cm  and  iocra. 
Find  the  circumference  lying  midway  between  them. 

30.  If  the  diameters  of  the  ends  of  a  smooth  log  are  3dm  and  5dm,  what  is 
the  diameter  of  a  section  midway  between  the  ends  ? 

31.  Find  the  exterior  and  interior  circumferences  of  a  circular  tank,  the 
exterior  diameter  being  i.o8m  and  the  thickness  of  the  side  3cm. 

32.  A  circular  reservoir  is  dug  g.6m  in  diameter,  j.2m  deep.     It  is  then 
lined  with  stones  28cm  long,  I2cm  wide,  8cm  high.     Find  the  number  of  stones 
required,  allowing  icm,  outside  measurement,  between  the  stones  for  mortar. 

33.  An  iron  wheel  1.26™  in  diameter  has  teeth  upon  its  circumference, 
distant  6cm  from  centre  to  centre  ;    (z.)  how  many  teeth  are  there,  and  («'.) 
what,  is  the  length  of  the  circumference  in  which  the  highest  points  of  the 
teeth  lie,  the  height  of  the  teeth  being  7cm? 

34.  If  the  length  of  the  minute-hand  of  a  watch  =i8mm,  what  distance 
does  its  end  pass  over  in  one  day  ?  in  one  year  ? 

35.  What  distance  is  traversed  by  a  point  in  the  circumference  of  a  water- 
wheel  2m  in  diameter,  if  the  wheel  makes  1400  revolutions  in  one  hour  ? 

36.  The  earth  turns  on  its  axis  once  a  day.     If  its  diameter  =  8000  miles, 
find  the  velocity  per  second  of  a  point  on  the  Equator. 

37.  What  ought  to  be   the  diameter  of  a  millstone  which  is  to  make 
100  revolutions  per  minute,  if  experience  teaches  that  to  secure  good  grind- 
ing a  point  on  the  circumference  should  have  a  velocity  of  about  8m  per 
second  ? 

38.  Draw  two  concentric  circles  with  the  radii  25cm  and  4Ocm;   then  de- 
scribe a  concentric  circle  whose  circumference  equals  the  sum  of  the  two  other 
circumferences.     Find,  also,  its  length. 

39.  What  is  the  radius  of  a  circle  whose  circumference  is  equal  to  the 
difference  between  the  circumferences  of  circles  24°™  and  45cm  in  diameter  ? 

40.  Find  the  diameter  of  a  circle  whose  circumference  is  equal  to  five 
times  that  of  a  circle  3Ocm  in  diameter. 

41.  Describe  two  circles,  making  the  circumference  of  one  equal  to  one- 
quarter  that  of  the  other. 


§   211.]  CHAPTER  IX.  —  THE   CIRCLE.  229 

42.  The  circumferences  of  two  concentric  circles  are  50™  and  8om.     How 
far  apart  are  they  ? 

43.  Of  two  toothed  wheels  which  work  together,  the  first  has  fifty,  the 
second  eighty  teeth.     How  many  revolutions  will  the  first  make  while  the 
second  makes  sixty  ?     What  diameter  has  the  first,  if  that  of  the  second  = 
1.32™? 

Hint.  —  The  distance  between  the  centres  of  the  teeth  on  both  wheels  must  be 
the  same. 

44.  The  diameter  of  a  toothed  wheel  having  100  teeth  =  6ocm.      How 
many  teeth  must  a  wheel  have,  which,  when  connected  with  the  first  wheel, 
makes  ten  revolutions  while  the  first  makes  one  ?      And  what  must  be  its 
diameter  ? 

45.  Of  two  belted  wheels  in  a  machine,  the  first  makes  twelve  revolutions 
while  the  second  makes  five.     If  the  diameter  of  the  first  wheel  —  g6cm,  find 
that  of  the  second. 

46.  Invent  and  describe  an  arrangement  of  toothed  wheels,  such  that, 
if  the  first  wheel   revolves  once  a  day,  the  last  wheel  will  revolve  once  a 
second. 

47.  The  hind  wheels  of  a  carriage  are  i.5m  in  diameter.      How  many 
revolutions  will  they  make  while  the  fore  wheels,  im  in  diameter,  make  2500 
revolutions  ? 

48.  In  a  circle  2m  in  diameter  find  the  length  of  the  arc  subtended  by 
one  side  of,  — 

(z.)     The  inscribed  square. 
(«.)    The  inscribed  regular  pentagon, 
(m.)  The  inscribed  regular  octagon. 
(zV.)  The  inscribed  regular  decagon. 

49.  Diameter  of  a  circle  =  3-3m.     How  long  is  an  arc  of  140°  ? 

50.  What  is  the  diameter  of  the  circle  if  an  arc  of  40°  has  a  length 
of  8.3m  ? 

51.  What  is  the  radius  of  a  circle  in  which  each  degree  of  the  circumfer- 
ence has  a  length  of  icm  ? 

52.  \Vhat  is  the  length  of  i°  on  a  protractor,  the  diameter  of  which  =  48cm? 

53.  Find  the  ratio  between  the  lengths  of  degrees  on  two  concentric  cir- 
cles, the  radii  of  which  are  idm  and  8dm. 

54.  How  many  degrees,  etc.,  are  there  in  an  arc  equal  in  length  to  the 
radius  of  the  circle  ? 

55.  Find  the  ratio  between  an  arc  of  a  quadrant  and  the  corresponding 
chord. 


230  GEOMETRY   FOR   BEGINNERS.  [§   212. 

VII.  —  Area,  of  the  Circle. 

§  212.  The  area  of  a  circle  is  always  less  than  that  of  a  circum- 
scribed regular  polygon,  and  greater  than  that  of  an  inscribed 
regular  polygon ;  moreover,  the  areas  of  these  polygons  approach 
nearer  to  each  other,  and  to  the  area  of  the  circle,  as  the  number 
of  their  sides  is  increased.  We  might,  therefore,  find  the  area  of 
a  circle  by  a  process  similar  to  that  employed  for  finding  the 
length  of  the  circumference.  The  following  method,  however,  is 
much  better :  — 

Circumscribe  a  regular  polygon  about  a  circle  (as  in  Fig.  204], 
and  let  r  =  its  less  radius  =  radius  of  the  circle.  Now  it  is  clear, 
that  by  increasing  the  number  of  sides  in  the  polygon,  we  can  make 
its  area  approach  the  area  of  the  circle  as  nearly  as  we  please,  but 
we  cannot  make  it  absolutely  equal  to  the  area  of  the  circle ;  in 

other  words,  — 

rthat  value  which  the  area  of  the  polygon 
Area  of  the  circle  =  -<  approaches,  but  never  absolutely  reaches,  as 

v.  the  number  of  its  sides  is  increased. 

This  value  of  the  area  of  the  polygon  is  termed  its  limiting  value, 
or  simply  its  LIMIT  ;  so  that  we  may  write  the  above  equation  more 
briefly,  thus  :  — 

Area  of  the  circle  =  limit  of  area  of  the  polygon. 

By  §  133,  area  of  the  polygon  =  J  r  x  its  perimeter.  This  equa- 
tion is  true,  no  matter  how  great  or  how  small  the  number  of  sides. 
Now,  by  increasing  the  number  of  sides,  the  perimeter  can  be  made 
to  approach  the  circumference  of  the  circle  as  nearly  as  we  please, 
but  can  never  quite  reach  it.  Therefore,  — 

Limit  of  area  of  the  polygon  =  %r  X  circumference  of  the  circle. 
Therefore  (Axiom  I.),  — 

Area  of  circle  =  J  r  X  its  circumference. 


§   214.]  CHAPTER   IX.  -  THE    CIRCLE.  231 


§  213.    If  we  substitute  for  the  circumference  the  values 
2  TT  r,  we  obtain  the  very  useful  formula,  — 

Area  of  a  circle  =  1  ird2  =  irr2.  [n.] 

For  example  :  if  the  radius  /-=  8m,  the  area  6"=  3.14  X  64  = 
200.96**. 

From  Formula  [n]  we  find  that  the  values  of  d  and  of  r  in 
terms  of  £  and  -  are,  — 


Read  these  formulas  in  words. 

F  20  /  -  ' 

For  example  :  if  S  =  2Oqm,  then  r  =  A  -  =  V6.37  =  2.52™. 

\3-i4 

In  Formula  [n]  assume  any  value  for  r  (or  d),  then  double  it, 
treble  it,  quadruple  it,  etc.,  and  work  out  for  each  case  the  cor- 
responding value  of  the  area  of  the  circle.  We  find  that  the  areas 
are  to  each  other  as  the  numbers  i,  4,  9,  16,  etc.  That  is,  — 

The  areas  of  circles  are  to  each  other  as  the  squares  of  their  radii 
(or  their  diameters}  . 

§  214.    Problem.  —  To  transform  a  given  circle  into  a  square. 

By  computation.  Find  the  area  of  the  circle,  extract  its  square 
root,  and  the  result  will  be  one  side  of  the  square  required. 

Let  a  =  the  radius  of  the  given  circle  ;  then  its  area  =  ~  a2,  and 
\he  side  of  the  equivalent  square  =  V^^2  =  a  V~  =  a  Vs.  14159 
6=3x1.77245. 

By  construction.  Draw  a  straight  line  equal  to  half  the  circum- 
ference, and  then  find,  by  §  170,  a  mean  proportional  between 
this  line  and  the  radius  ;  this  will  be  a  side  of  the  required  square. 

Remark.  —  A  square  which  shall  be  exactly  equivalent  to  a 
given  circle  cannot  be  found,  because  the  value  of  TT  cannot  be 
exactly  found.  In  other  words,  the  problem  of  "  squaring  the 
circle  "  admits  only  of  an  approximate  solution.  But  the  value 


232 


GEOMETRY    FOR    BEGINNERS. 


[§215- 


^  =  3.14159  gives  a  result  far  more  accurate  than  could  be  actually 
constructed  with  ruler  and  compasses. 

Exercises.  —  1.  Draw  a  circle,  and  then  find  the  side  of  an  equivalent 
square,  (z.)  by  computation,  (z'z.)  by  construction.  How  do  the  results 
agree  ? 

2*    Construct  a  square,  and  then  make  a  circle  equivalent  to  the  square. 
3.    Find  the  ratio  of  the  areas  of  a  square  and  a  circle,  if,  — 

(z.)    A  side  of  the  square  is  equal  to  the  radius  of  the  circle. 
(zY.)  A  side  of  the  square  is  equal  to  the  diameter  of  the  circle. 


215.  Upon  the  sides  of  the  right  triangle  ABC  (Fig.  208), 
as  diameters,  describe  circles.  By 
§142,— 


If  we  multiply  the  terms  of  this 
equation  by  £TT,  the  equality  is  not 
destroyed  (Axiom  IV.),  and  we  ob- 
tain, — 

£TT  x  TC2=  i-xZ#24-  J*  X  ~BC\ 
Now  these  three  quantities,  taken 
in  order,  are  the  areas  of  the  circles 
described  upon  the  hypotenuse  and 
the  two  legs  as  diameters. 

Theorem.  —  The  area  of  the  circle  described  upon  the  hypote- 
nuse of  a  right  triangle  as  diameter  is  equal  to  the  sum  of  the  areas 
of  the  circles  described  upon  the  two  legs  as  diameters. 

NOTE.— This  is  a  special  case  of  a  very  general  theorem,  which  may  be  called 
the  Generalized  Theorem  of  Pythagoras,  namely  :  if  upon  the  sides  of  a  right  trian- 
gle we  construct  any  three  similar  figures,  the  area  of  the  figure  constructed  upon  the 
hypotenuse  will  be  equal  to  the  sum  of  the  areas  of  the  figures  con 


Fig.  208. 


legs. 


constructed  upon  the 


Exercises.  —  1.    Make  a  circle  equal  to  the  sum  of  two  given  circles. 

2.  Make  a  circle  equal  to  twice  a  given  circle. 

3.  Make  a  circle  equal  to  the  difference  of  two  given  circles. 

4.  Make  a  circle  equal  to  half  of  a  given  circle. 


§     2  1 6.] 


CHAPTER    IX. THE    CIRCLE. 


233 


A 


§  216.  Problem.  —  To  find  the  area  of  a  circular  sector  AOB 
(Fig.  209],  having  given  the  radius  of  the  circle  and  the  angle  of 
the  sector. 

Solution.  —  From  §  182,  Corollary,  and 
§  212.  it  follows  that  the  area  of  a  sector 
is  found  by  multiplying  its  arc  by  half  the 
radius  of  the  circle. 

Example.  —  Let  the  radius  =  14™,  and 
the  angle  of  the  sector  =  40°.  Then,  — 

Circumference  of  circle  =  2  x  3!  X  14. 

Arc  of  sector  =  ^°Ty  X  circumference. 

Area  of  sector  =  its  arc  x  7  =  68.44qm. 

Compute  the  sector  for  the  following 
values  of  the  angle  :  90°,  180°,  270°. 

§  217.  Problem.  —  71?  find  the  area  of  a  circular  segment 
C ED  (Fig.  209),  having  given  the  radius  of  the  circle -,  the  angle 
at  the  centre  COD,  and  the  height  E F  of  the  segment. 

Solution.  —  Compute  the  areas  of  the  sector  COD  and  the  tri- 
angle COD.  Then,  segment  C  ED  =  sector  COD  —  triangle 
COD. 

Example.  —  Let  the  radius  =  i™,  the  angle  COD  =  40°,  and 

the  height  EF=  6cm.     Then  (§  216),  the  sector  COD  =  ^L 

360 

=  -  =  o.349qm.     In  the  triangle  COD,  the  altitude  OF  =  OE  — 

9  

EF=  0.94™  ;  by  §  187,  I.,  half  the  base  CF  =  \lEF(2  -  EF) 
=  0.342™ ;  the  base  CD  =  0.684™ ;  and  the  area  of  the  triangle 
=  °'94  X  0.684  =  0.32I48qm.  Therefore,  the  area  of  the  seg- 
ment =  0.349  —  0.32148  =  o.o2752qm  :  an  answer  correct  enough 
for  all  practical  purposes. 

NOTE.  — The  height  E  F  (Fig.  209}  of  a  segment,  and  the  angle  COD  at  the 
centre,  are  not  independent  of  each  other,  but  are  connected  by  a  relation  which 
it  is  the  business  of  Trigonometry  to  investigate. 


234  GEOMETRY  FOR  BEGINNERS.  [§  2 1 8. 

§  218.    PRACTICAL  EXERCISES. 

(  Take  TT  =  3^,  unless  otherwise  stated.) 

Find  the  area  of  a  circle,  having  given,  — 

1.  The  diameter  =  I  im.  4.    The  radius  = 

2.  The  diameter  =  4.37™.  5.    The  radius  = 

3.  The  diameter  =  9™.  0.    The  radius  =  4|milcs. 
Find  the  area  of  a  circle  (TT  =  3.14159),  having  given,— 

7.  The  diameter  =  8.346™.  9.   The  radius  =  27.19™. 

8.  The  diameter  =  i6.37km.  10.    The  radius  =  5 ift. 
Find  the  radius  of  a  circle,  having  given,  — 

11.  The  area  =  100^.  14.   The  area  =  2ooosi-  <*. 

12.  The  area  =  946<im.  15.   The  area  =  682.4«i- in. 

13.  The  area  =  23.37^.  16.    The  area  =  72^- miles. 

17.  The  radius  of  a  circle  =  7™.     Find  the  area,  («'.)  taking  TT=  3},  («.) 
7r=3-I4I59-     What  is  the  difference  (in  square  centimeters)  of  the  results? 
Which  result  is  more  correct? 

18.  The  diameter  of  a  circular  park  is  ikm.     Find  the  difference  in  its 
areas  computed,  (j.)  taking  TT  —  3!;    (».)  taking  TT  =  3.14159. 

19.  A  man  buys  a  circular  field,  diameter  =  i6ooft,  for  four  cents  per  square 
foot.     What  difference  will  there  be  in  the  price  paid  for  the  field  according 
as  TT  is  taken  equal  to  3^  or  equal  to  3.14159? 

20.  Describe  a  circle,  and  then  find  its  area. 

21.  Find  the  area  of  a  circle  if  its  circumference  =  8om. 

22.  The  circumference  of  a  tree  measures  2.6m.    Find  the  area  of  a  cross- 
section. 

23.  If  the  length  of  a  circular  race-course  is  to  be  imile,  how  many  acres 
of  land  will  it  cover? 

24.  If  the  circumference  of  a  garden-bed  =  48.25m,  what  is  its  area? 

25.  Around  the  banks  of  a  circular  pond  are  planted  256  poplars,  6m  apart. 
Find  the  area  of  the  pond. 

26.  What  will  it  cost  to  cover  a  round  table  the  radius  of  which  =  75°™, 
W  Sie  cloth  is  6ocm  wide  and  costs  $2.00  per  meter? 

Find  the  radius,  the  circumference,  and  the  area  of  a  circle  equal  to,  — 

27.  The  sum  of  two  circles  whose  diameters  are  5™  and  6m. 

28.  The  sum  of  two  circles  whose  areas  are  6oim  and  8ofim. 

29.  The  difference  of  two  circles  whose  radii  are  3m  and  4™. 

*JO»    The  difference  of  two  circles  whose  circumferences  are  33™  and  44™. 


§  2l8.]  CHAPTER   IX.  —  THE   CIRCLE.  235 

Find  the  radius  of  a  circle, — 

31.  Three  times  as  large  as  a  circle  whose  radius  =  7m. 

32.  Five  and  one-half  times  as  large  as  a  circle  whose  diameter  =  I2.6m. 

33.  One-ninth  as  large  as  a  circle  whose  circumference  =  45.23™. 

34:.  If  a  horse,  tied  by  a  rope  4m  long  to  a  stake  driven  in  the  ground, 
eats  all  the  grass  which  he  can  reach  in  three  hours,  how  long  must  the  rope 
be  in  order  that  the  grass  may  last  twelve  hours? 

35.  If  the  diameters  of  the  holes  in  the  side  of  a  tank  are  as  3  :  I,  how 
much  more  water  will  flow  in  an  hour  out  of  one  than  out  of  the  other? 

36.  If  80  flowers  will  grow  on  a  circular  garden-bed,  and  I  wish  another 
bed  on  which  800  of  the  same  flowers  will  grow,  what  diameter  must  it  have 
compared  with  that  of  the  first  bed? 

37.  Find  the  area  of  a  circular  ring  (Fig.  209},  the  radii  of  the  circles 
being  6™  and  7™. 

38.  Find  the  area  of  a  ring,  the  two  circumferences  being  2Om  and  30™. 

39.  Describe  two  concentric  circles;   then  find  the  area  of  the  circular  ring. 

40.  In  the  middle  of  a  circular  pond  400™  in  diameter  stands  a  circular 
island  90™  in  diameter.     Find  the  area  of  the  surface  covered  by  the  water. 

41.  Find  the  area  of  a  ring  if  the  outer  circumference  =  85cm  and  the 
breadth  of  the  ring  =  iocm. 

42.  What  will  it  cost  to  have  a  walk  im  wide  about  a  circular  reservoir 
64™  in  diameter,  the  price  of  paving  being  $2.50  per  square  meter? 

43.  A  target  consists  of  a  black  circle  o.75m  in  diameter,  surrounded  by  a 
white  ring  0.28™  wide.     Find  the  total  area  covered  by  the  target. 

44.  The  outer  and  inner  circumferences  of  a  round  tower  are  17.2™  and 
I2.8m.     Find  the  ground  area  covered  by  the  wall. 

45.  How  wide  is  a  ring  if  the  areas  of  the  two  circles  are  2OOim  and  641™? 

46.  If  a  ring  is  to  be  4m  wide,  and  is  to  contain  Scfl™,  what  must  be  the 
radii  of  the  two  circles? 

47.  To  the  shorter  sides  of  a  table,  2m  by  1.25'",  semicircular  leaves  are 
added.     Find  the  total  area  of  the  table. 

48.  Find  the  total  area  of  the  figure  formed  by  adding  semicircles  to  the 
four  sides  of  a  square,  one  side  of  which  =  i.2m. 

49.  A  square,  a  rectangle,  and  a  circle  have  the  same  perimeter,  8m.    Find 
the  three  areas,  the  rectangle  being  two-thirds  as  wide  as  it  is  long. 

50.  The  circumference  of  a  cifcle  is  equal  to  the  perimeter  of  a  square 
whose  side  is  iom.     Find  the  difference  in  their  areas. 

NOTE.  —  Of  all  plane  figures  with  equal  perimeters  the  circle  has  the  greatest 
area. 


236  GEOMETRY    FOR    BEGINNERS.  [§   2l8. 

51.  A  circle  and  a  square  have  the  same  area,  i2Cflm.     Find  the  difference 
between  the  circumference  of  the  circle  and  the  perimeter  of  the  square. 

52.  If  a  circle,  a  square,  a  regular  hexagon,  and  a  regular  octagon  each 
contain  io.4a,  find  the  differences  between  the  perimeters  of  the  polygons  and 
the  circumference  of  the  circle. 

53.  A  circular  piece  of  lead  is  recast  in  the  shape  of  a  square,  the  thick- 
ness remaining  the  same.     If  the  radius  of  the  circle  =  45cm,  find  a  side  of  the 
square. 

54.  Find  the  diameter  of  a  circle  equivalent  to  a  regular  octagon  whose 
side  =  2m. 

55.  The  radius  of  a  circle  =  8m.    Find  the  difference  between  its  area  and 
the  area  of  the  inscribed  regular  hexagon. 

56.  In  a  picture-frame  6ocm  square  is  a  circular  picture,  the  radius  of  which 
=  2Ocm.     Find  the  area  of  the  part  not  covered  by  the  picture. 

57.  In  a  circle  whose  radius  =  3Ocm,  find  the  area  of  a  sector  of  80°. 

58.  If  the  radius  of  a  circle  =  76°™,  find  the  area  of  a  sector  whose  arc  is 
2Ocm  long. 

59.  How  large  is  a  sector  if  its  arc  contains  120°  and  is  40°™  long? 

60.  What  part  of  the  entire  circle  is  the  sector  whose  arc  equals  the  radius 
of  the  circle? 

61.  If  the  area  of  a  sector  of  45°  =  2£im,  find  the  radius  of  the  circle. 

62.  The  radius  of  a  circle  =  3™,  the  height  of  a  segment  =  i.968m,  and  the 
angle  of  the  segment  =  98°.     Find  the  area  of  the  segment. 

In  a  circle  whose  radius  —  4m,  find  the  area  of  the  segment  cut  off  by  the 
side  of, — 

63.  The  inscribed  equilateral  triangle. 
G4.    The  inscribed  square. 

65.  The  inscribed  regular  hexagon. 

66.  The  inscribed  regular  octagon. 


CHAPTER    IX. REVIEW.  237 

REVIEW   OF   CHAPTER   IX. 

SYNOPSIS.1 

I.  — Sectors,  Angles  at  the  Centre,  Chords,  Segments. 

§  180.  Review  of  §§  26-28.  Definitions  of  a  sector,  an  angle  at  the  centre,  a 
chord,  a  diameter,  a  semi-circumference,  greater  and  less  arcs,  a  seg- 
ment, greater  and  less  segments.  Six  corollaries. 

§  181.  Equal  angles  at  the  centre,  their  arcs,  chords,  sectors,  and  segments. 
Five  theorems.  Four  corollaries.  Definition  of  a  quadrant, 

§  182.  Unequal  angles  at  the  centre,  their  arcs  and  sectors.  One  theorem. 
One  corollary. 

§  183.  Properties  of  a  diameter  perpendicular  to  a  chord.  Two  theorems. 
One  problem.  Two  corollaries. 

§  184.  Equal  and  unequal  chords,  their  angles  at  the  centre,  and  their  dis- 
tances from  the  centre.  Four  theorems. 

II. —  Inscribed  Angles. 
§  185.   Definition  of  an  inscribed  angle. 

§  186.   Measure  of  an  inscribed  angle.     One  theorem.     Four  corollaries. 
§  187.   Application  of  §  166  to  the  circle.     Two  theorems, 
§  188.    Product  of  the  parts  of  a  chord  passing  through  a  point  within  a 
circle.     One  theorem.     One  corollary. 

III.  —  Secants  and  Tangents. 

§  189.    Definitions  of  a  secant  and  a  tangent. 

§  190.  Relation  between  a  tangent  and  a  radius  drawn  to  the  point  of  con- 
tact. One  theorem. 

§  191.  Tangent  to  a  circle  through  a  given  point.  One  problem.  One  corol- 
lary. 

§  192.   Measure  of  the  angle  between  a  tangent  and  chord.     One  theorem. 

IV.  — Two  Circles. 

§  193.    Definition  of  concentric  and  eccentric  circles,  a  ring,  the  line  of  centres. 
§  194.    Relative  positions  of  two  eccentric  circles.     Five  cases. 


1  Here,  and  in  the  remaining  chapters,  the  synopsis  is  presented  in  a  condensed 
form  which  may  be  made  the  basis  of  a  more  or  less  extended  review  according 
to  the  discretion  of  the  teacher. 


238  GEOMETRY    FOR    BEGINNERS. 

V.  —  Inscribed  and  Circumscribed  Figures. 

§  195.   Definitions  of  inscribed  and  circumscribed  figures. 

§  196.   Circle  inscribed  in  a  triangle.     One  corollary. 

§  197.    Circle  circumscribed  about  a  triangle.     One  corollary. 

§  198.  Relation  between  the  sides  of  a  circumscribed  quadrilateral.  One 
theorem.  One  corollary. 

§  199.  Relation  between  the  angles  of  an  inscribed  quadrilateral.  One  theo- 
rem. One  corollary. 

§  200.    Circle  inscribed  in  a  regular  polygon.     One  theorem. 

§  201.    Circle  circumscribed  about  a  regular  polygon.     One  theorem. 

§  202.   Division  of  a  circumference  into  equal  parts.    Three  cases. 

§  203.    Regular  polygon  inscribed  in  a  circle. 

§  204.    Regular  polygon  circumscribed  about  a  circle. 

§  205.   Construction  of  regular  polygons. 

§  206.  An  inscribed  regular  polygon,  and  the  inscribed  regular  polygon  with 
twice  as  many  sides. 

§  207.  An  inscribed  regular  polygon,  and  the  circumscribed  regular  polygon, 
with  the  same  number  of  sides. 

VI.  —  Length  of  a  Circumference. 

§  208.  The  general  truths  which  enable  us  to  find  the  length  of  a  circumfer- 
ence. Method  of  applying  these  truths.  Value  of  TT. 

§  209.  Formula  for  the  length  of  a  circumference.  Relation  between  two 
circumferences,  their  radii,  and  their  diameters. 

§  210.   Two  ways  of  estimating  the  length  of  an  arc. 

§  211.   Practical  exercises. 

VII.  — Area  of  a  Circle. 
§  212.    Method  of  finding  the  area  of  a  circle. 
§  213.   Formula  for  the  area  of  a  circle.   Relation  between  the  areas  of  circles, 

their  radii,  and  their  diameters. 
§  214.   A  circle  and  the  equivalent  square. 
§  215.   Relation  between  the  areas  of  circles  described  upon  the  sides  of  a 

right  triangle  as  diameters. 
§  216.   Area  of  a  sector. 
§  217.  Area  of  a  segment. 
§  2ia   Practical  exercises. 


CHAPTER    IX. REVIEW.  239 


EXERCISES. 

1.  In  a  circle,  as  the  angle  at  the  centre  increases,  the  corresponding  arc, 
chord,  sector,  and  segment  also  increase.     Which  increase  at  the  same 
rate  as  the  angle,  and  which  do  not  ? 

2.  Upon  a  given  straight  line  as  a  chord  describe  a  segment  which  shall 
contain  a  given  angle. 

Hints. —  If  AB  is  the  given  line,  draw  A  C,  making  B  A  C  equal  the  given 
angle.  Then  find  the  centre  of  the  circle  which  has  AB  for  a  chord,  and 
which  A  C  touches  at  A.  Then  apply  §§  189  and  192. 

3.  Construct  a  right  triangle,  having  given  the  hypotenuse  and  a  line  in 
which  the  vertex  of  the  right  angle  must  lie. 

4.  In  a  given  circle  draw  a  chord  parallel  to  a  given  straight  line  and  having 
a  given  length. 

5.  Find  the  locus  of  all  points  from  which  tangents  drawn  to  a  given  circle 
shall  have  a  given  length  (§§73,  190). 

6.  Given  two  circles  ;    find  the  point  from  which  tangents  to  the  circles 
shall  have  the  same  length. 

7.  Draw  a  circle  which  shall  touch  two  given  parallel  lines  and  pass  through 
a  given  point  lying  between  the  lines.     (Two  solutions.) 

8.  About  a  given  point  describe  a  circle  that  shall  touch  a  given  circle. 

(The  point  may  be  either  without  or  within  the  given  circle.) 

9.  About  each  of  two  given  points  describe  a  circle  so  that  the  two  circles 
shall  touch  each  other,  and  one  of  them  shall  touch  a  given  straight  line. 

Hint.  —  First  construct  the  circle  which  touches  the  line,  and  then  use  the 
preceding  exercise. 

10.  With  a  given  radius  describe  a  circle  which  shall  touch  a  given  circle  and 
pass  through  a  given  point.     (Discuss  the  problem.) 

11.  Construct  a  triangle,  having  given  two  sides  and  the  radius  of  the  cir- 
cumscribed circle. 

12.  Construct  a  triangle,  having  given  two  angles  and  the  radius  of  the  in- 
scribed circle. 

13.  Divide  a  given  circle  into  four  equal  parts  by  describing  concentric  circles. 

14.  If  the  perimeter  of  a  sector  is  equal  to  the  circumference  of  the  circle, 
find  the  angle  at  the  centre. 

15.  If  the  area  of  a  sector  is  equal  to  the  square  of  the  radius  of  the  circle, 
find  the  angle  at  the  centre. 

16.  If  the  perimeter  of  a  sector  is  equal  to  twice  the  diameter  of  the  circle, 
find  the  ratio  of  the  sector  to  the  circle. 


240  GEOMETRY   FOR   BEGINNERS.  [§  219. 

CHAPTER  X. 
THE    ELLIPSE. 

CONTENTS.  — I.   The  Properties  of  the  Ellipse  (§§  219-221).     II.  Construction  oi 
Ellipses  (§§  222-224). 

I.— The  Properties  of  the  Ellipse. 

§  219.   Among  curved  lines  the  ellipse  ranks  in  importance  next 
to  the  circle. 

Fig.  210  shows  how  an  ellipse  may  be  described.     Fasten  a 

piece  of  paper  upon  a  smooth  sur- 
face, stick  in  two  pins  at  any  dis- 
tance apart,  and  tie  to  the  pins  the 
ends  of  a  string  longer  than  the 
distance  apart  of  the  pins.  Then 
describe  the  curve  with  a  pencil,  as 
shown  in  the  figure,  taking  care  to 
keep  the  string  constantly  stretched 
to  its  full  length. 

As  the  point  of  the  pencil  moves 
Fig.  210.  around  the  pins,  its  distance  from 

either  one  of  the  pins  is  constantly 

changing ;  but  it  is  clear  that  the  sum  of  its  distances  from  the  two 
pins  must  always  remain  the  same. 

Thus,  in  Fig.  210,  AP  +  BP  =  AQ  +  BQ  =  AR  -f  B R,  etc. 
Definition.  —  The  ELLIPSE  is  a  curve  such  that  the  sum  of  the 
distances  of  every  point  in  it  from  two  fixed  points  is  the  same. 
The  two  fixed  points  are  called  the  foci. 
The  point  O,  half-way  between  the  foci,  is  called  the  centre. 
Define  the  ellipse  regarded  as  the  locus  of  a  moving  point. 


§    220.] 


CHAPTER    X. — THE    ELLIPSE. 


241 


The  distance  OA  or  OB  from  the  centre  to  either  of  the  foci,  is 
called  the  eccentricity  of  the  ellipse. 

The  less  the  eccentricity,  —  that  is  to  say,  the  nearer  the  foci  are 
to  each  other,  —  the  wider  the  ellipse  becomes  in  proportion  to  its 
length,  and  the  more  nearly  it  approaches  to  the  form  of  the  circle. 

Describe  several  ellipses  with  their  foci  at  different  distances  from 
.each  other. 

What  does  the  ellipse  become  if  the  foci  coincide,  or  what  comes 
to  the  same  thing,  if  only  one  pin  is  used,  to  which  both  ends  of 
the  thread  are  fastened? 

NOTE  i. — The  ellipse  is  a  more  beautiful  curve  than  the  circle.  Ellipses,  or 
curves  closely  resembling  ellipses,  are  much  used  in  the  arts;  in  cabinet  work 
(tables,  mirrors,  etc.),  in  various  designs  and  patterns,  in  the  arches  of  bridges,  and 
in  architecture.  It  is  to  the  skilful  use  of  the  ellipse  and  still  more  complex  curves, that 
the  superiority  of  Greek  art  over  that  of  other  nations  is  largely  due.  Astronomy 
presents  us  with  examples  of  ellipses  on  the  grandest  scale.  Our  earth  and  all 
the  planets  of  the  Solar  System  revolve  around  the  sun  in  elliptical  paths  in  obe- 
dience to  the  law  of  gravitation,  the  sun  being  at  one  of  the  foci. 

NOTE  2.  —  The  ellipse  is  connected  with  the  circle  in  a  remarkable  way.  A 
section  of  a  cylinder  (page  4,  Fig.  3)  made  by  a  plane  parallel  to  the  bases  is  (like 
the  bases)  a  circle ;  but  if  the  section  is  inclined  to  the  bases,  the  boundary  of  the 
section  is  an  ellipse;  and  the  more  inclined  the  section,  the  longer  the  ellipse  in 
proportion  to  its  breadth.  All  this  is  easily  shown  by  making  sections  through  a 
wooden  cylinder. 

If  we  look  at  a  circle  on  paper  with  one  eye  placed  directly  in  front  of  the 
centre  of  the  circle,  and  then  gradually  turn  the  paper  till  its  edge  comes  in  front  of 
the  eye,  we  shall  find  that,  as  the  paper  is  turned  around,  the  circle  will  present  the 
appearance  of  an  ellipse  which  becomes  narrower  and  narrower,  and  tends  to  pass 
finally  into  a  straight  line.  In  this  way  we  pass,  by  imperceptible  gradations,  from 
the  circle  to  the  straight  line. 

§  220.  In  an  ellipse  CDEF  (Fig.  211),  the  line  CD,  which 
passes  through  the  foci,  and  is  lim- 
ited by  the  ellipse,  is  called  the 
Major  Axis ;  and  the  line  E  F, 
which  passes  through  the  centre  O 
perpendicular  to  the  major  axis, 
and  is  limited  by  the  ellipse,  is 
called  the  Minor  Axis. 

I.    From   the   definition   of  the 
ellipse,  it  follows   that  AC  +  BC  =  A  D  +  B  D,  or  2  AC  -f 


242  GEOMETRY    FOR    BEGINNERS.  [§   221. 

AB  =  2  B  D  -f  A  B  j  whence  (Axiom  III.)  2  A  C  =  2  23 D,  and 
AC=  BD;  that  is,  - 

The  vertices  of  an  ellipse  are  at  the  same  distance  from  the  foci. 

II.  Since  OA=OB,  and  AC  =  BD;  therefore  (Axiom  II.), 
OA  +  AC=  OB  +  BD,  or  OC  =  OD;  that  is,  - 

The  centre  of  an  ellipse  bisects  the  major  axis. 

III.  Because  (Fig.  210),  AP+  BP  =  AC  +  BC,  and  ^C  = 
^Z>;  therefore,  ^/>  +  BP=BD  +  BC  =  CD-,  that  is,— 

77/<?  sum  of  the  distances  of  any  point  of  the  ellipse  from  the  foci 
is  equal  to  the  major  axis. 

IV.  Since  A  A  O  E  ^  A  B  O  E,  therefore  A  E  =  B  E ;  and  since 
AE  -f  BE  =  CD,  therefore  2AE^  CD,  and  AE  =  £  CZ>  = 
BE  ;  that  is,  — 

The  ends  of  the  minor  axis  are  equidistant  from  the  foci,  the 
distance  being  equal  to  half  the  major  axis. 

V.  Since  A  A OE  ^/^AOF,  therefore  OE  =  OF ;  that  is,  — 
The  centre  of  an  ellipse  bisects  the  minor  axis. 

Exercises.  —  1.  To  what  are  the  three  sides  of  the  right  triangle  AOE 
{Fig.  2ii\  respectively  equal  ? 

2.  What  kind  of  plane  figure  \&AEBF  (Fig.  211}  ?     Why  ? 

3.  Given  the  axes  of  an  ellipse,  find  the  positions  of  the. foci. 

4.  Given  the  major  axis  and  the  positions  of  the  foci,  find  the  minor  axis. 

5.  Semi-major  axis  =  18™;   semi-minor  axis=  I2m.     Find  the  eccentricity. 

6.  Semi-major  axis  =  1. 3m;  eccentricity  =  0.34™.   Find  the  semi-minor  axis. 
1.    Semi-minor  axis=  2.9™;  eccentricity  =  0.9™.    Find  the  semi-major  axis. 
8.    Major  axis  =  58dm;   minor  axis  =  45dm.     How  far  apart  are  the  foci  ? 
9»   The  path  of  the  earth  about  the  sun  (at  one  focus)  is  an  ellipse  whose 

axes  are  20,657,70x3  and  20,655,100  geographical  miles.     How  can  the  least 
and  the  greatest  distances  of  the  earth  from  the  sun  be  computed  ? 

§  221.  AREA  OF  AN  ELLIPSE.  —  If  we  describe  circles  about  the 
axes  of  an  ellipse  as  diameters  (see  Fig.  213),  it  is  obvious  that 
the  area  of  the  ellipse  is  less  than  that  of  the  circle  with  the  semi- 
major  axis  as  radius,  and  greater  than  that  of  the  circle  with  the 


§   221.]  CHAPTER    X. THE    ELLIPSE.  243 

semi-minor  axis  as  radius.  Now  it  can  be  proved  that  the  area  of 
the  ellipse  is  exactly  equal  to  that  of  the  circle  whose  radius  is  a 
mean  proportional  between  the  semi-axes  of  the  ellipse. 

If  a  and  b  are  the  semi-axes  of  an  ellipse,  r  the  radius  of  the 
circle  equal  in  area  to  the  ellipse,  then  a  :  r  —  r :  b,  or  r2  —  a  b. 
The  area  of  this  circle  =  -r2  (§  213) ;  therefore  the  area  of  the 
ellipse  =  TC  r2 ;  or,  since  r2  =  a  b, 

Area  of  the  ellipse  =  irab.  [12.] 

That  is  to  say,  the  area  of  an  ellipse  is  equal  to  the  continued 
product  of  its  semi-axes  and  the  number  TT. 

Exercises.  —  Find  the  area  of  an  ellipse,  if,  — 

1.  The  semi-axes  are  I2dm  and  8dm.     Find  the  area. 

2.  The  axes  are  8.5™  and  4.5™.     Find  the  area. 

3.  The  major  axis  =  9™,  the  eccentricity  =  1.5™.     Find  the  area. 

4.  The  minor  axis  =  2m,  the  eccentricity  =  im.     Find  the  area. 

5.  The  major  axis  =  5™,  the  focal  distance  =  3™.     Find  the  area. 

6.  The  area=  681™,  the  major  axis  =  I2m.     Find  the  minor  axis  and  the 
eccentricity. 

7.  How  much  cloth  will  cover  an  elliptical  table  2.4™  long,  i.3m  wide  ? 

8.  The  famous  amphitheatre  at  Verona,  built  by  the  Emperor  Domitian, 
and  which  seated  24,000  spectators,  has  for  its  base  an  ellipse  whose  axes  are 
133™  and  105™.     Find  the  area  of  this  ellipse. 

9.  The  section  of  an  arch  has  the  shape  of  a  semi-ellipse  i6m  long,  4.8m 
high.     Find  the  area  of  the  section. 

10.  The  diameter  of  a  circle  =  7™.     Upon  this  diameter  as  major  axis  an 
ellipse  is  constructed  half  as  large  as  the  circle.     Find  its  minor  axis. 

11.  A  pond  in  the  shape  of  an  ellipse,  with  the  axes  6om  and  44™,  is  changed 
to  a  circle  with  the  radius  6om.     How  much  larger  is  it  than  before  ? 

12.  Find  the  radius  of  a  circle  equal  in  area  to  an  ellipse  whose  axes  are 
I3m  and  9™. 

13.  Find  the  side  of  a  square  equal  in  area  to  an  ellipse  whose  axes  are 
I7m  and  iom. 

14.  Find  the  area  of  the  largest  ellipse  that  can  be  made  out  of  a  board 
2m  long  and  o.6m  wide. 

15.  The  axes  of  an  ellipse  are  4m  and  3™.     Upon  these  axes  as  diameters 
circles  are  described.     Find  the  areas  of  all  three  figures. 


244  GEOMETRY    FOR    BEGINNERS.  [§   222. 

II.  —  Construction  of  Ellipses. 

§  222.  Problem.  —  To  construct  an  ellipse,  having  given  the 
major  axis  and  the  eccentricity. 

Let  {Fig.  212}  CD  be  the  given  major  axis,  O  its  middle  point. 
3^3  Make  OA  =  OB  —  the  given  ec- 

centricity ;  then  A  and  B  are  the 
foci.  Find  the  points  E  and  F, 
}D  such  .  that  the  distance  from  each 
of  them  to  A  and  to  B  is  equal 
to  the  semi-major  axis ;  E  and 
F  are  the  ends  of  the  minor 
axis. 

In  order  to  find  more  points  of  the  ellipse,  assume  any  points 
m,  11,  p,  etc.,  lying  in  the  major  axis  between  A  and  O.  With  a 
radius  equal  to  Cm,  describe  arcs  from  each  focus  as  centre,  both 
above  and  below  CD\  and  likewise  do  the  same  with  a  radius 
equal  to  Dm  ;  the  four  points  marked  i,  where  these  arcs  intersect, 
are  four  points  of  the  ellipse.  For,  by  this  construction,  the  sum 
of  the  lines  drawn  from  either  of  the  points  marked  i  to  the  two 
foci  is  equal  to  the  major  axis.  (See  §  220,  III.) 

In  like  manner,  by  the  use  of  n,  the  points  marked  2,  and  by  the 
use  of/,  those  marked  3,  are  found. 

By  drawing  (free-hand)  through  all  these  points  a  curved  line, 
making  the  curvature  as  nearly  uniform  as  possible,  we  obtain  a 
curve  which  approaches  the  nearer  to  the  exact  ellipse  required, 
the  greater  the  number  of  points  that  are  previously  determined. 

Exercises.  —  1.  How  can  an  ellipse  be  described  by  continuous  motion 
(§  219).  (Gardeners  sometimes  use  this  method  to  trace  the  outline  of  an 
elliptical  flower-bed.) 

2»  Construct  the  ellipse  whose  major  axis  =  8ocm  and  eccentricity  =  3Ocm, 
by  finding  sixteen  points  in  the  ellipse. 

i1*   Construct  the  ellipse  whose  major  axis  =  6ocm  and  eccentricity  =  i6cm, 
fourteen  points. 


§    223.] 


CHAPTER    X. THE    ELLIPSE. 


245 


§  223.    Problem.  —  To  construct  an  ellipse,  having  given  the 
axes,  by  the  aid  of  circles  described  about  the  axes  as  diameters. 

Let  AB  and  CD  {Fig.  213)  be  the  given  axes  intersecting  at  O. 
Describe  circles  about  A  B  and  CD 
as  diameters.  Through  O  draw  any 
radii  O  E,  OF,  etc.  Through  the 
points  where  these  radii  cut  the 
inner  circle  draw  lines  II  AB,  and 
through  the  points  where  the  radii 
cut  the  outer  circle  draw  lines  II 
CD.  The  intersection  of  each 
pair  of  these  lines  is  a  point  in  the 
required  ellipse ;  so  that  nothing 
remains  but  to  draw  (free-hand) 
through  all  the  points  of  intersection 
a  smooth  curve. 


Fig.  213. 


Exercises.  —  1.  Construct  an  ellipse  having  for  its  axes  72°™  and  4Ocm,  and 
find  the  positions  of  the  foci. 

2.  Construct  an  ellipse  whose  major  axis  is  to  the  minor  axis  as  5  :  4,  and 
find  the  positions  of  the  foci. 


§  224.    Problem.  —  To  construct  with  arcs  of  circles  a  curve 
that  resembles  an  ellipse. 

Draw  a  line  A  D  (Fig.  214) ,  and  divide  it  into  three  equal  parts, 
AB,  BC,  CD.     With  a  radius  equal  to 
one  of  these  parts  describe  about  B  and 
C  as  centres  two  circles,  cutting  each 
other   in   E    and   F.      Through    these  A\ 
points    and  the   two   centres   draw  the 
lines  EG,  EH,  FI,  FK.     Then   de- 
scribe about  E  and  F  as  centres,  with 
a  radius  equal  to  EG,  the  arcs 


Fig.  214. 


IK.     The  curve  AGPID IK  is  the  required  curve. 


246 


GEOMETRY    FOR    BEGINNERS. 


[§   224. 


The  elliptical  curves  employed  in  the  arts  are  usually  constructed 
in  this  or  a  similar  way  from  arcs  of  circles.  By  compounding 
arcs  of  circles,  in  different  ways,  a  great  variety  of  curves  may  be 
constructed.  Some  examples  are  added  below,  in  Figs.  215—222. 
Explain  how  each  figure  is  constructed. 


Fig.  215. 


Fig.  216. 


Fig.  217. 


Fig.  218. 


Fig.  219. 


Fig.  220. 


§    224/! 


CHAPTER   X. THE    ELLIPSE. 


247 


Fig.  22T. 


Fig.  222. 


Exercises.  —  1.    Construct  an  elliptical  qirve  with  arcs  of  circles. 

2.  Construct  the  curve  shown  in  Fig.  215. 

3.  Construct  the  curve  shown  in  Fig.  216. 

4.  Construct  the  curve  shown  in  Fig.  217. 

5.  Construct  the  curve  shown  in  Fig.  218. 

6.  Construct  the  curve  shown  in  Fig.  219. 

7.  Construct  the  curve  shown  in  Fig.  220. 

8.  Construct  an  aval  like  that  in  Fig.  221. 

9.  Construct  a  spiral  like  that  in  Fig.  222. 

10.    Find  the  area  of  an  oval  like  that  of  Fig.  221,  if  the  diameter  AB  =  4™. 
Explain  each  step  of  the  solution,  stating  the  truth  on  which  it  depends. 


SYNOPSIS  OF  CHAPTER  X. 

§  219.   Definition  of  the  ellipse,  its  foci,  its  centre,  its  eccentricity.     Relation 

of  the  ellipse  to  the  circle. 
§  220.   Definition  of  the  axes  of  an  ellipse.     Relations  between  the  ends  of 

the  axes,  the  foci,  and  the  centre.     Five  general  relations. 
§  221.   Area  of  an  ellipse.     Formula  for  the  area. 

§  222.   Construction  of  an  ellipse,  given  the  minor  axis  and  the  eccentricity. 
§  223.   Construction  of  an  ellipse  with  the  aid  of  circles,  given  the  axes  of  the 

ellipse. 
§  224.   Construction  of  a  curve  resembling  an  ellipse  by  means  of  arcs  of 

circles.     Other  instances  of  the  use  of  circular  arcs  in  constructing 

curves. 


248  GEOMETRY    FOR    BEGINNERS.  [§   225. 


CHAPTER   XI. 
PLANES. 

CONTENTS.  —  I.  Straight  Lines  in  Space  (§§  225-227).  II.  A  Plane  (§§  228-230). 
III.  A  Plane  and  a  Straight  Line  ($$  231-235).  IV.  Two  Planes  ($$  236-239). 
V.  Three  Planes  ($  240).  VI.  Solid  Angles  (§  241). 

I.  —  Straight  Lines  in  Space. 

§  225.  Thus  far  the  figures  studied  (triangles,  polygons,  circles, 
etc.)  are  such  as  can  be  represented  in  their  true  shape  on  paper 
or  the  blackboard ;  they  are  figures  which  lie  wholly  in  one  plane. 

We  shall  now  proceed  to  consider  some  of  the  properties  of 
figures  which  are  not  confined  to  one  plane.  Every  actual  body 
which  we  can  see  or  handle  is  an  example  of  such  a  figure. 

We  shall  begin  by  examining  the  relations  of  lines  and  planes  in 
space  to  one  another. 

NOTE  i. —  In  this  part  of  Geometry  i\\e  first  and  cAief  difficulty  encountered  by 
beginners  lies  in  conceiving  and  holding  firmly  in  the  mind  the  true  relative  posi- 
tions of  the  lines  and  planes  under  consideration.  Without  the  power  to  do  this 
the  proofs  of  theorems  amount  to  mere  words  ;  with  it  the  proofs  are  easily  mastered  ; 
in  fact  many  of  the  theorems  will  seem  self-evident.  Accordingly,  in  the  brief  space 
here  devoted  to  the  subject,  rigorous  proofs  will  not  always  be  given,  but  care  will 
always  be  taken  to  set  forth  the  truth  so  that  it  shall  be  completely  realized  in 
thought.  The  geometric  imagination  must  be  exercised  and  strengthened  before 
the  mind  is  prepared  to  study  the  subject  with  all  the  rigor  of  logic. 

NOTE  2.  —  An  additional  difficulty  arises  from  the  fact  that  in  representing 
magnitudes  in  different  planes  upon  a  single  plane  surface  (that  of  paper  or  the 
blackboard)  these  magnitudes  must,  in  general,  be  changed  in  size  or  shape  or 
both.  Lines  and  angles  must  be  altered  in  size,  plane  figures  in  both  size  and 
shape.  The  rules  for  doing  this  correctly  are  based  on  the  principles  of  Geometry, 
and  constitute  the  Art  of  Perspective,  of  Isometric  Drawing,  etc.  There  is  one 
general  rule  of  Isometric  Drawing  which  it  is  useful  to  bear  in  mind  :  a  square,  not 
in  the  plane  of  the  drawing,  is  represented  in  that  plane  by  a  rhombus,  a  rectangle 
by  a  parallelogram,  and  a  circle  by  an  ellipse.  Beginners  should  make  a  habit  of 
comparing  the  figures  in  the  book,  and  those  which  they  draw  themselves,  with  actual 
lines  and  planes  placed  in  the  proper  relative  positions,  until  they  are  able  to  recog- 
nize at  once  the  true  relative  positions  of  the  parts  of  a  drawing,  and  to  make  similar 
drawings  themselves.  For  lines,  wires  and  stretched  threads  may  be  employed; 
for  planes,  paper,  cardboard,  a  thin  piece  of  board,  glass,  the  floor  and  sides  of  the 


§  226.]  CHAPTER   XI. PLANES.  249 

room ;  for  bodies,  suitable  models  made  of  wood  or  cardboard  (see  figs.  1-6)  or 
constructed  simply  of  stiff  wire  for  the  edges. 

NOTE  3.  —  Some  of  the  relations  of  lines  and  planes  have  been  noticed  in  Chap- 
ter I.  and  in  §§  24,  25.  The  learner  should  refresh  his  memory,  if  necessary,  by 
reviewing  them. 

§  226.  Two  straight  lines  in  space  may  have  either  of  three 
relative  positions,  as  follows  :  — 

(/.)     They  may  be  parallel  to  each  other. 

(//.)    They  may  intersect  each  other. 

(///.)  They  may  be  neither  parallel  nor  intersecting. 

In  the  first  two  cases  the  lines  must  lie  in  the  same  plane  (§  29). 
Give  examples  of  both  cases.  If  the  lines  intersect,  they  may  be 
either  perpendicular  to  each  other  or  inclined  to  each  other  (§  48). 

In  the  third  case  the  two  lines  do  not  lie  in  the  same  plane.  A 
line  from  north  to  south  on  the  table,  and  a  line  from  east  to  west 
on  the  floor,  is  an  example  of  two  such  lines. 

Let  AB  and  CD  (Fig.  22j)  represent  two  lines  which  are  not 
parallel  and  do  not  intersect,  and  let  E 
be  the  point  of  A  B  which  is  nearest  to 
CD,  and  F  the  point  of  CD  which  is 
nearest  to  AB.  Then  the  line  E  F  is 
the  shortest  distance  between  the  lines  ABA  E  .  B 

and  CD,  and  is  a  common  perpendicular  Fig,  223. 

to  both  lines. 

NOTE.  —  In  Fig.  223,  A  B  is  to  be  regarded  as  in  the  plane  of  the  paper,  CD  as 
nearly  perpendicular  to  this  plane. 

§  227.  Let  ABC  and  DEF  (Fig.  224)  be  two  angles  in  space, 
and  let  AB  II  DF,  and  BC  II  F  F. 
Conceive  the  angle  ABC  to  move 
so  that  its  vertex  B  keeps  in  the  line 
BE,  and  AB  and  CB  to  remain 
parallel  to  their  first  positions.  When 
B  reaches  F,  it  is  evident  that  A  B 
will  coincide  with  DE,  and  CB  with 


250  GEOMETRY    FOR    BEGINNERS.  [§   228. 

FE,  and  therefore  the  angle  ABC  will  coincide  with  the  angle 
DEF. 

Therefore,  the  theorem  of  Plane  Geometry  (§  58)  that  angles, 
whose  sides  are  respectively  parallel  and  directed  the  same  way 
from  the  vertex,  are  equal,  also  holds  true  of  any  two  angles  .in 
space. 

II.  —  A  Plane. 

§  228.  The  surface  of  still  water,  the  surface  of  a  table,  the  floor, 
or  the  blackboard,  are  examples  of  plane  surfaces,  or  planes.  Give 
another  example.  How  can  a  plane  be  generated  by  motion? 
(See  §  28.) 

How  was  a  plane  denned  in  §  28  ?  And  what  is  the  test  of  a 
plane  surface? 

The  following  is  the  more  abstract  but  more  precise  definition 
commonly  given  by  mathematicians  :  — 

Definition.  —  A  PLANE  is  a  stirface  such  that  the  straight  line 
which  joins  any  two  points  in  the  surface  lies  wholly  in  the  surface. 

A  plane  (like  a  straight  line)  may  be  conceived  to  extend  indefi- 
nitely. The  surface  of  the  table  may  be  extended  in  thought  as 
far  as  we  please,  and  in  such  a  way  that  the  surface  which  we 
imagine  will  be  everywhere  a  part  of  the  same  plane  as  the  actual 
surface  which  we  see. 

Exercise.  —  Give  an  instance  of  a  vertical  plane,  a  horizontal  plane,  an 
inclined  plane. 

§  229.  When  we  conceive  a  plane  as  containing  a  given  point 
or  straight  line,  we  are  said  to  pass  the  plane  through  the  point  or 
the  line. 

Through  one  point  an  endless  number  of  planes  in  all  conceiva- 
ble positions  may  be  passed. 

Through  two  points,  also,  we  may  pass  as  many  different  planes 


§   230.]  CHAPTER    XI. PLANES.  251 

as  we  please ;  for  if  we  pass  a  plane  through  the  line  joining  the 
points  A  and  B  (Fig.  22 5) ,  and  then  rotate  the  plane  about  this 
line  as  an  axis,  it  will  assume  different  positions,  in  all  of  which  it 
contains  the  line,  and  therefore  the  points  A  and  B. 

But  if  a  third  point  C  be  given,  through  which  the  plane  must 
also  pass,  it  is  evident  that  as  we  rotate  the  plane  there  is  only  one 
position  in  which  it  will  contain  all  three  points.  Through  three 
points  not  in  the  same  straight  line  only  one  plane  can  be  passed ; 
and  the  same  is  obviously  true  of  a  straight  line  and  a  point  not  in 
the  line. 

It  is  likewise  true  of  two  intersecting  lines  and  of  two  parallel 
lines  ;  for  if  we  pass  a  plane  through  one  of  the  lines,  and  then  turn 
the  plane  about  this  line,  there  is  only  one  position  in  which  it  can 
contain  the  other  line. 

Therefore,  a  plane  is  determined,  — 

1 .  By  three  points  not  in  the  same  straight  line. 

2.  By  a  straight  line  and  a  point  not  in  the  line. 

3.  By  two  intersecting  lines. 

4.  By  two  parallel  lines. 

§  230.   Any  two  planes  which  are  passed  through  a  straight  line 
AB  (Fig.  2 25}  have  this  line  for  their 
intersection ;  and  it  is  quite  clear  that 
through  a  curved  line  only  one  plane 
at  most  can  be  passed. 

The  intersection   of  two  planes  is 
always  a  straight  line.  i  ^^^ 

Exercises.  —  1.    How  many  planes  can  Fig.  225. 

intersect  in  the  same  straight  line  ? 

2.  How  many  planes  can  be  passed  through  a  curve  drawn  on  the  black- 
board ? 

3.  Make  (with  a  wire)  a  curve  such  that  no  plane  can  be  passed  through  it 


252  GEOMETRY    FOR    BEGINNERS.  [§   231. 

III.— A  Plane  and  a  Straight  Line. 

§  231.    A  straight  line  may  be  either,  — 

(/.)     Parallel  to  a  plane. 

(«'.)    Perpendicular  to  a  plane. 

(Hi.)  Inclined  to  a  plane. 

An  edge  of  the  ceiling  is  parallel  to  the  floor ;  a  plumb-line  is 
perpendicular  to  the  surface  of  still  water ;  a  rafter  of  a  common 
roof  is  inclined  to  the  level  ground.  Give  other  examples. 

In  Fig.  226,  the  line  AB  is  parallel  to  the  plane  M.1  The  line 
PO  is  perpendicular  to  Mt  and  the  lines  PC  and  PD  are  inclined 
to  M. 

Exercises. — 1.  Draw  a  figure  like  Fig.  226,  to  illustrate  the  different 
positions  which  a  straight  line  may  have  relative  to  a  plane. 

2.  Hold  a  pencil,  (z.)  parallel,  («.)  perpendicular,  (iii)  inclined,  to  the 
table. 

§  232.  Definition.  —  A  straight  line  and  a  plane  are  parallel, 
if  they  cannot  meet  however  far  extended. 

This  condition  will  be  fulfilled  if  the  straight  line  A  B  (Fig.  226) 

is  parallel  to  a  straight  line  C E  in 
the  plane ;  for,  since  A  B  can  never 
meet  CE,  and  can  never  leave  the 
plane  determined  by  AB  and  CJl 
(§  229,4),  it  can  never  meet  the  plane 
M  in  which  C  E  lies.  Hence,  — 

I.    A  straight  line  is  parallel  to  a 
p-    22fr  plane  if  it  is  parallel  to  a  straight 

line  drawn  in  the  plane. 

II.  If  two  lines  are  parallel,  every  plane  passed  through  one  of 
the  lines  will  be  parallel  to  the  other  line. 

What  exception  is  there  to  this  last  statement  ? 

1  In  naming  a  plane,  a  single  letter,  as  M  or  N,  is  usually  sufficient. 


§  233-] 


CHAPTER   XL  —  PLANES. 


253 


Exercises.  —  1.  How  many  lines  parallel  to  a  plane  can  be  drawn  through 
a  point  outside  the  plane  ? 

2.  If  two  lines  are  parallel  to  a  plane,  are  they  parallel  to  each  other  ? 
Give  illustrations. 

§  233.  A  straight  line,  perpendicular  or  inclined  to  a  plane,  will 
(prolonged,  if  necessary)  meet  the  plane  in  a  point.  This  point 
is  called  the  foot  of  the  line. 

Definition.  —  A  straight  line  is  perpendicular  to  a  plane,  if  it 
is  perpendicular  to  every  straight  line  that  can  be  drawn  through 
its  foot  in  the  plane. 

It  is  also  said  to  be  normal  to  the  plane. 

A  line  having  this  property  can  always  be  drawn  from  a  point 
P  (Fig.  227)  to  a  plane  M.  For, 
among  all  the  lines  that  can  be 
drawn  from  P  to  meet  the  plane, 
there  will  be  one,  PO,  shorter  than 
any  other.  If  we  draw  through  its 
foot  O  any  straight  lines  AB,  CD, 
etc.,  PO  will  be  the  shortest  line 
from  P  to  each  of  these  lines  ;  there- 
fore (§  83),  perpendicular  to  each 
and  all  of  them. 

Hence,  a  perpendicular  measures  the  distance  from  a  point  to  a 
plane. 

Through  the  intersection  O  (Fig.  228)  of  two  lines,  AB  and  CD, 
draw  any  line  O  E  _L  CD,  and  revolve  OE  about  CD,  keeping 
OE  _L  CD ;  there  is  one,  and  only 
one,  position  in  which  the  line  will 
also  be  J_  AB.  Let  OP  be  this 
position ;  then  O  P,  being  the  only 
line  through  O  perpendicular  to  both 
A  B  and  CD,  must  be  perpendicular 
to  the  plane  passed  through  A  B  and 
CD.  Therefore,  — 


227. 


Fig.  228. 


254  GEOMETRY    FOR    BEGINNERS.  [§   234. 

If  a  straight  line  is  perpendicular  to  two  straight  lines  drawn 
through  a  point  of  the  line,  it  is  perpendicular  to  the  plane  contain- 
ing those  two  lines. 

This  proposition  enables  us  to  solve  the  problem  : 
To  make  a  plane  perpendicular  to  a  straight  line. 
We  have  only  to  draw  through  any  point  of  the  line 
two  perpendiculars,  and  then  to  pas:;  a  plane  through 
these  perpendiculars.  Explain  how  a  carpenter  would 
proceed  in  order  to  saw  squarely  in  two  the  beam  of 
wood  shown  in  Fig.  229. 

Exercises.  —  1.    Give  examples  of  a  line  perpendicular  to 

aPlane' 

2.    When  a  right  triangle  revolves  about  one  of  its  legs, 

what  is  generated  by  the  other  leg  ? 

3.  At  a  point  of  a  straight  line  in  space,  how  many  perpendiculars  can  be 
erected  ?     From  a  point  not  in  the  line,  how  many  perpendiculars  can  be  let 
fall  to  the  line  ? 

4.  Find  the  distance  from  a  point  P  to  a  plane,  if  its  distance  from  a 
point  A  in  the  plane  =  a,  and  the  distance  of  this  point  from  the  foot  of  the 
perpendicular  let  fall  from  the  first  point  to  the  plane  =  b.     Solve  for  the 
case  where  a  =  17,  b  =  8. 

5.  Prove  that  lines  drawn  from  a  point  meeting  a  plane  at  equal  distances 
from  the  foot  of  the  perpendicular  let  fall  from  the  point  are  equal ;    and  that 
of  two  lines  drawn  to  unequal  distances  from  the  foot  of  the  perpendicular, 
the  more  remote  is  the  greater. 

6.  Prove  the  converse  of  the  first  part  of  the  last  exercise. 

7.  Find  the  locus  of  points  in  a  plane  equidistant  from  points  not  in  the 
plane. 

8.  Find  the  locus  of  points  equidistant  from  two  given  points. 

9.  Find  the  locus  of  points  equidistant  from  three  given  points. 


§  234.  Let  A  B  and  CD  (Fig.  236)  be  normal  to  the  plane  J/, 
and  suppose  AB  to  move  along  AC,  keeping  parallel  1  )  its  first 
position.  Then  its  position  relative  to  the  plane  will  not  change, 
that  is,  it  will  remain  normal  to  the  plane  ;  and' when  A  coincides 


§  '350 


CHAPTER   XI. PLANES. 


255 


with    C,  AB  will  coincide  with    CD.     Therefore,  AB   II    CD. 
That  is,— 

Two  lines  that  are  normal  to  a  plane  are  parallel  to  each  other. 

The  converse  is  also  true.     State  it. 

Of 


Fig.  2jo. 


Fig.  231. 


§  235.  Lines  neither  parallel  nor  perpendicular  to  a  plane  are 
inclined  to  the  plane. 

Let  A  O  (Fig.  2ji)  be  inclined  to  the  plane  M.  Draw  AB  _L  M, 
and  join  OB ;  then  OB  is  called  the  PROJECTION  of  the  line  AO 
upon  the  plane.  When  is  the  projection  of  a  line  equal  in  length 
to  the  line  ?  less  than  the  line  ?  When  is  it  a  point  ? 

The  acute  angle  A  OB,  which  A  O  makes  with  its  projection  BO, 
is  the  smallest  angle  which  A  O  makes  with  any  line  drawn  through 
its  foot  on  the  plane  M.  To  see  this  more  clearly,  revolve  BO 
about  O  in  the  plane  M.  The  angle  which  the  revolving  line 
makes  with  AO  will  continually  increase,  becoming  equal  to  a 
right  angle  AOC  after  one-fourth  of  a  revolution,  and  attaining 
its  greatest  obtuse  value  AOD  after  half  a  revolution.  During  the 
remaining  half  of  the  revolution  what  changes  does  the  angle 
undergo  ? 

The  angle  which  a  line  inclined  to  a  plane  makes  with  its  pro- 
jection in  the  plane,  is  called  the  inclination  of  the  line  to  the 
plane. 

NOTE.  —  What  is  said  above  may  be  illustrated  by  drawing  lines  through  a 
point  on  a  sheet  of  paper,  and  placing  the  end  of  the  pencil  held  in  an  inclined 
position  against  this  point.  If  in  this  position  we  let  the  pencil  drop,  the  angle 
which  it  describes  in  falling  is  the  inclination  of  the  pencil  to  the  table. 


256 


GEOMETRY    FOR    BEGINNERS. 


[§  236- 


IV.—  Two  Planes. 

§  236.   Two  planes  may  be  either,  — 

(/.)     Parallel  to  each  other/ 

(«'.)    Perpendicular  to  each  other. 

(«V.)  Inclined  to  each  other. 

The  floor  and  ceiling  of  the  room  are  parallel  planes ;  the  floor 
and  a  side  of  the  room  are  perpendicular  to  each  other ;  the  floor 
and  the  roof  of  the  house  are  inclined  to  each  other.  Give  other 
examples. 

Exercises.  —  1.  In  Fig.  232,  which  planes  are  parallel?  perpendicular? 
inclined  ? 

2»    Draw  two  planes  of  each  kind. 

3.  Hold  two  planes  (books,  thin  sticks  of  wood,  etc.)  parallel ;  perpen- 
dular ;  inclined. 


Fig.  232. 


Fig-  233. 


§  237.  Definition.  —  Two  planes  are  parallel  if  they  cannot 
meet  however  far  extended. 

This  condition  is  fulfilled  if  the  planes  M  and  N  (Fig.  2jj)  are 
both  perpendicular  to  the  same  line  AB.  For  if  we  pass  a  third 
plane  through  AB,  it  will  intersect  M  in  a  line  CD  \-AB  (why?), 
and  N  in  a  line  EF ^LAB.  Therefore,  CD  and  EFt  being  in 
the  same  plane,  are  parallel  (§  57),  and  can  never  meet.  The 
same  reasoning  may  be  applied  to  all  planes  that  can  be  passed 
through  AB ;  therefore,  the  planes  J/and  TV  can  never  meet. 


§   238.]  CHAPTER   XI. — PLANES.  257 

Therefore,  — 

I.  Planes  that  have  a  common  normal  are  parallel. 
The  converse  is  also  true.     State  it. 

It  is  also  evident  from  what  precedes,  that,  — 

II.  The  intersections  of  two  parallel  planes  by  a  third  plane  are 
parallel  lines. 

If,  also,  CE  II  DF  (Fig.  233),  then  CDFE  is  a  parallelogram 
(§  101),  and  AC  =  BD  (§  102) ;  that  is,  - 

III.  Parallel  lines  between  parallel  planes  are  equal. 

Among  these  equal  parallel  lines  are  the  common  normals 
(§  234),  which  measure  the  distance  apart  of  the  planes;  there- 
fore, — 

IV.  Parallel  planes  are  everywhere  equally  distant. 

§  238.  Definition.  —  Two  planes  are  perpendicular  to  each 
other  if  a  normal  to  either  plane,  erected  at  any  point  of  their 
intersection,  lies  in  the  other  plane. 

The  angle  made  by  the  normals  to  the  two  planes,  erected  at  the 
same  point  of  the  intersection,  is  a  right  angle  (why  ?) . 

Let  the  planes  M  and  N  (Fig.  234}  intersect  in  the  line  AB, 
and  let  CD  be  normal  to  M  and  lie  in  N. 

Draw  D  E  normal  to  N\  then  C  D  E  is  a  right  angle  (why?)  ; 
therefore    DE    must  lie   in  M,  the 
plane    containing    all    lines    perpen- 
dicular to  CD  at  D. 

Further,  let  G  be  any  other  point 
of  the  intersection,  and  draw  GF  nor- 
mal to  My  and  G  H  normal  to  N.  Then 
GF  II  DC  (§  234) ;  and  since  DC 
lies  in  the  plane  N,  and  G  also  lies  in 
N,  therefore  GF  must  lie  in  N.  For 
like  reasons,  GH  lies  in  M. 

Comparing  these  results  with  the  definition,  we  see  that,  — 


258  GEOMETRY   FOR    BEGINNERS.  [§    239. 

1.  Two  planes  must  be  perpendicular,  if  a  single  normal  to  one  of 
the  planes  erected  at  a  point  of  the  intersection  lies  in  tJie  other  plane. 

II.  Every  plane  passed  through  a  normal  to  a  plane  is  perpen- 
dicular to  that  plane. 

Exercises.  —  1,  How  can  a  plane  perpendicular  to  a  given  plane  be  passed 
through  a  given  point  ? 

2.  How  can  a  plane  perpendicular  to  a  given  plane  be  passed  through  a 
straight  line,   (z.)   when  the  line   lies  in  the   given  plane  ?    (zY.)   when  it  is 
inclined  to  the  given  plane  ?    (zYY.)  when  it  is  parallel  to  the  given  plane  ? 

§  239.  Two  inclined  planes  M  and  N  (Fig.  235}  intersect  each 
other  and  form  what  is  called  a  DIHEDRAL  ANGLED  This  angle 

may  be  conceived  as  generated  by 
the  revolution  of  one  of  the  planes  M 
about  the  line  A  B  until  it  comes  into 
the  position  of  the  other  plane  N. 
The  planes  M  and  N  are  called  the 
faces  of  the  angle ;  the  intersection 

rAB  is  called  the  edge  of  the  angle. 
A  dihedral  angle  is  always  measured 
by  the  angle  of  two  lines.    If  OC  is  J_ 

A~        ~~~  AB,  during  the  revolution  of  M  about 

AB,  OC  describes  the  angle  COD, 

which  has  the  same  value  wherever  the  point  O  is  situated  in  the 
line  AB  (§  227).  This  angle  may  therefore  be  used  to  measure 
the  dihedral  angle  made  by  the  planes.  That  is,  — 

A  dihedral  angle  is  measured  by  the  angle  made  by  two  lines 
drawn  through  any  point  of  the  edge,  perpendicular  to  the  edge,  one 
line  in  one  plane,  the  other  in  the  other  plane. 

Exercises. — 1.  Illustrate,  by  the  leaves  and  edges  of  a  book,  dihedral 
angles,  and  how  they  are  measured. 

2.  If  the  angle  of  the  lines  which  measure  a  dihedral  angle  is  90°,  what 
position  do  the  planes  have  ? 

1  From  Greek  words  meaning  two-faced. 


§  240.] 


CHAPTER   XI. PLANES. 


259 


V.  — Three  Planes. 

§  240.   Three  planes  may  have  five  different  relative  positions, 
shown  in  Figs.  236-238. 

(/.)   In  Fig.  236,  the  planes  are  parallel  to  one  another. 


7 


V 


Fig.  236. 


Fig.  237. 


(«.)  In  Fig.  237,  the  two  planes  J/and  N  are  parallel,  and  the 
third  plane  P  (or  Q)  intersects  M  and  N  in  parallel  lines.  (See 
§  237.) 

(///.)  In  Fig.  237,  the  planes  N,  P,  and  Q  have  a  common  line 
of  intersection. 

(iv.)  In  /3g-.  2J7,  J/,  P,  and  <?  intersect  in  three  lines,  which 
are  parallel  to  one  another. 


Fig.  238. 


Fig.  239. 


.)  In  Fig.  238  there  are  also  three  lines  of  intersection.  These 
lines  meet  in  a  common  point  O  j  so  that  O  is  a  point  common  to 
all  the  planes. 


260  GEOMETRY    FOR    BEGINNERS.  [§   241. 

If  two  of  the  planes  J/and  N  (Fig.  239)  are  both  perpendicu- 
lar to  the  third  plane  R,  it  follows,  from  §  238,  that  their  inter- 
section must  be  a  normal  to  the  plane  R ;  for  the  normal  to  R, 
erected  at  the  point  O,  must  lie  in  both  M  and  N, 

Exercise.  —  Give  an  example  of  three  parallel  planes  ;  or'  two  planes,  both 
perpendicular  to  a  third  plane. 

VI.  — Solid  Angles. 

§  241.  When  three  or  more  planes  meet  in  one  point  O  (Fig. 
241)  they  are  said  to  form  at  that  point  a  SOLID  ANGLE.  The 
planes  are  called  the  faces,  their  lines  of  in- 
tersection the  edges,  and  the  angles  at  (9, 
made  by  the  two  edges  of  each  plane,  the 
plane  angles  of  the  solid  angle. 

If  there  are  only  three  planes,  the  angle 
is  usually  called  a  trihedral  angle. 

In  order  to  form  a  trihedral  angle  from 
three  angles  in  one  plane,  AO B  =  a,  AOC 
=  b,  BOD  —  c  (Fig.  241),  of  which  we 
will  suppose  a  to  be  the  greatest,  turn  the 
plane  AOC  about  A  O,  and  the  plane  AOD 

about  BO,  till  they  meet,  if  possible,  and  the  lines  CO  and  DO 
coincide. 

If  b  -f-  c  =  a,  CO  and  DO  will  just  coincide  when  turned  through 
half  a  revolution,  so  that  both  again  lie  in  the  plane  of  the  paper ; 
if  b  +  f<a,  CO  and  DO  cannot  be  made  to  coincide.  In  both 
cases  no  trihedral  angle  will  be  formed.  But  if  b  -f-  c  >  a,  CO  and 
DO  will  coincide  in  a  line  OF,  either  in  front  of  the  plane  of  the 
paper  (Fig.  242),  or  behind  it  (Fig.  243),  according  as  the  motion 
is  forward  or  backward ';  and  a  trihedral  angle  is  formed.  And, 
since  a  is  supposed  the  greatest  of  the  three  angles,  it  follows  that 
a  +  b  >  c,  and  that  a  -f-  c  >  b.  Hence,  in  every  trihedral  ang  -he 
sum  of  any  two  of  the  plane  angles  is  greater  than  the  third. 


§ 


CHAPTER   XL PLANES. 


261 


If  the  motion  is  forwards,  the  trihedral  angle  shown  in  Fig.  242 
is  formed  ;  if  backwards,  the  trihedral  angle  shown  in  Fig.  243. 
These  two  angles  are  composed  of  equal  plane  angles,  taken  in 


Fig.  241. 


Fig.  242. 


Fig.  243. 


order ;  yet  they  cannot  be  made  to  coincide,  because  their  equal 
parts  follow  in  reverse  order,  —  in  one,  from  left  to  right;  in  the 
other,  from  right  to  left. 

They  stand  to  each  other  in  the  same  relation  as  an  object  to  its 
image  in  a  mirror,  or  as  the  right  hand  to  the  left.  Two  such  solid 
angles  are  said  to  be  symmetrical. 

Two  solid  angles  are  equal,  if  they  are  composed  of  equal  plane 
angles  taken  in  the  same  order.  We  may  conceive  two  equal 
solid  angles  placed  one  upon  the  other  so  as  to  coincide  in  all 
their  parts. 

Whatever  be  the  number  of  plane  angles  which  compose  a  solid 
angle,  their  sum,  estimated  in  degrees,  must  be  less  than  j6o°;  if 
it  were  equal  to  360°,  the  plane  angles  would  necessarily  lie  in  one 
plane,  and  would  not  form  a  solid  angle. 

Exercises.  —  1.    Give  examples  of  solid  angles. 

2.  In  a  solid  angle,  how  is  the  angle  which  two  faces  make  with  each 
other  measured? 

3.  Draw  a  solid  angle  having  three  faces ;   four  faces  ;   six  faces. 


262  GEOMETRY    FOR    BEGINNERS. 


REVIEW   OF  CHAPTER  XL 

I.  —  Straight  Lines  in  Space. 

§  225.   Nature  of  Solid  Geometry. 
§  226.   Two  straight  lines  in  space. 
§  227.  Two  angles  in  space. 

II.  —  A  Plane. 

§  228.    Definition  of  a  plane. 

§  229.    Determination  of  a  plane. 

§  230.    Intersection  of  two  planes. 

III.  — A  Straight  Line  and  a  Plane. 

§  231.    Positions  of  a  straight  line  with  respect  to  a  plane. 

§  232.   The  straight  line  parallel  to  a  plane. 

§  233.   The  straight  line  perpendicular  to  a  plane. 

§  234.   Two  normals  to  a  plane. 

§  235.   The  straight  line  inclined  to  a  plane. 

IV. —  Two  Planes. 

§  236.  Relative  positions  of  two  planes. 

§  237.  Two  parallel  planes. 

§  238.  Two  perpendicular  planes. 

§  239.  Two  inclined  planes. 

V.  — Three  Planes. 
§  240.    Relative  positions  of  three  planes. 

VI.  —  Solid  Angles. 

§  241.    Solid  angles.     Symmetrical  solid  angles.     Equal  solid  angles. 


§    242.']  CHAPTER   XII. GEOMETRICAL    BODIES.  263 

CHAPTER  XII. 
GEOMETRICAL     BODIES. 

CONTENTS.  —  I.   Polyhedrons  (§§  242,243).     II.  The  Prism,  Cylinder,  Pyramid, 
and  Cone  (§§  244-250).     III.   The  Sphere  (§$  251,  252). 

J.  —  Polyhedrons. 

§  242.  A  limited  portion  of  space  regarded  as  possessing  -solely 
form  and  magnitude  is  called  a  GEOMETRICAL  BODY,  or  a  SOLID 
(see  §  7). 

In  this  sense  a  room  is  a  body  as  much  as  a  stone  or  piece  of 
wood.  Speaking  exactly,  it  is  the  space  occupied  by  the  air,  etc., 
in  the  room,  and  limited  by  the  floor,  sides,  and  ceiling  of  the  room. 

A  geometrical  body  bounded  on  all  sides  by  planes  is  called  a 
POLYHEDRON. 

Give  examples  of  polyhedrons. 

The  bounding  planes  are  called  the  faces  of  the  polyhedron, 
their  intersections  are  called  the  edges,  and  the  intersections  of 
the  edges  are  called  the  corners. 

The  faces  that  meet  at  each  corner  form  a  solid  angle,  composed 
of  as  many  plane  angles  as  there  are  faces. 

A  polyhedron  must  have  at  least  four  faces ;  for  it  takes  at  least 
three  planes  to  form  a  solid  angle,  and  a  fourth  plane  is  required 
to  close  the  opening  between  the  three  planes. 

A  polyhedron  with  four  faces  is  called  a  TETRAHEDRON  ;  with  five 
faces,  a  PENTAHEDRON  ;  with  six  faces,  an  HEXAHEDRON  ;  with 
eight  faces,  an  OCTAHEDRON  ;  with  ten  faces,  a  DECAHEDRON  ;  with 
twelve  faces,  a  DODECAHEDRON  ;  with  twenty  faces,  an  ICOSAHEDRON. 

A  very  simple  relation  exists  between  the  faces,  corners,  and 
edges  of  a  polyhedron. 


264  GEOMETRY    FOR    BEGINNERS.  [§   243. 

If  we  examine  the  polyhedrons  in  Fig.  2,  page  4,  we  find  that,  — 
No.  I.  has  5  faces,  6  corners,  and  9  edges. 
II.  has  6     "       8       "          "   12      " 

III.  has  8     "     12       "          "   1 8      " 

IV.  has  5     "       5  "8      " 

In  each  of  these  cases,  the  number  of  faces  4-  number  of  cor- 
ners =  number  of  edges  +  2- 

This  relation  is  perfectly  general,  and  is  called  Euler's  Theorem.1 
Let  F  denote  the  number  of  faces,  C  the  number  of  corners,  E  the 
number  of  edges  ;  then  we  may  state  Euler's  Theorem  as  follows  :  — 

In  every  polyhedron,        F  +  C  =  E  +  2.  [13.] 

§  243.  A  polyhedron  whose  faces  are  equal  regular  polygons, 
and  whose  solid  angles  are  all  equal,  is  called  REGULAR. 

It  is  easy  to  show  that  only  five  regular  polyhedrons  are  possible. 

We  know  already  that  at  least  three  faces  are  necessary  to  form 
a  solid  angle  ;  also,  that  the  sum  of  the  plane  angles  at  each  corner 
must  be  less  than  four  right  angles  ( §  241) . 

The  angle  of  the  equilateral  triangle  =  60°,  and  3  X  60°  —  180°  ; 
therefore,  three  such  triangles  may  form  a  solid  angle.  Likewise, 
four  such  triangles  may  be  used,  since  4  X  60°  =  240° ;  also  five 
such  triangles,  since  5  X  60°  =  300°.  But  6  X  60°  =  360°,  so  that 
six,  or  more  than  six,  equilateral  triangles  cannot  form  a  solid  angle. 

Therefore,  only  three  regular  polyhedrons  are  possible  whose 
faces  are  equilateral  triangles. 

The  angle  of  a  square  =  90°.  Since  3  X  90°  =  270°,  while 
4  x  90°  =  360°,  three  is  the  only  number  of  squares  that  can 
form  a  solid  angle. 

The  angle  of  a  regular  pentagon  =  108°.  Since  3  x  108°  —  324°, 
while  4  x  1 08°  =  432°,  only  three  regular  pentagons  can  be  used 
to  form  a  solid  angle. 

1  Euler  was  a  famous  German  mathematician,  born  1707,  died  1783. 


§  2430 


CHAPTER   XII. GEOMETRICAL   BODIES. 


265 


The  angle  of  a  regular  hexagon  =  1 20°.  Since  3  X  120°  =  360°, 
three  hexagons  placed  together  with  a  corner  common  would  lie 
in  one  plane  and  could  not  form  a  solid  angle.  Therefore,  no 
solid  angle  can  be  formed  with  regular  hexagons ;  still  less  with 
regular  polygons  having  more  than  six  sides. 


Fig-  247- 

Therefore,  only  five  regular  polyhedrons  are  possible.     They 
are,— 

(1)  The   regular  Tetrahedron   (Fig.  244),   bounded    by  four 
equilateral  triangles. 

(2)  The    regular   Octahedron   (Fig.  245},  bounded   by  eight 
equilateral   triangles. 

(3)  The  regular  Icosahedron  (Fig.  246),  bounded  by  twenty 
equilateral  triangles. 

(4)  The   regular    Hexahedron    (Fig.    247},    bounded   by   six 
squares. 

(5)  The  regular  Dodecahedron  (Fig.  248),  bounded  by  twelve 
regular  pentagons. 


266 


GEOMETRY    FOR   BEGINNERS. 


[§   243- 


To  DEVELOP  1  the  surface  of  a  solid  is  to  draw  in  one  plane  the 
various  parts  of  the  surface,  so  that  when  folded  up  they  shall  form 
the  solid. 

The  surface  of  the  solid,  when  drawn  in  one  plane,  is  called  the 
DEVELOPMENT  of  the  surface. 

The  surfaces  of  many  solids  cannot  be  developed ;  but  when 
this  can  be  done,  it  is  easy  to  construct  by  this  means  models  of 
the  solids. 

The  surfaces  of  the  regular  polyhedrons  are  all  developable,  and 
their  plane  models  are  shown  in  Figs.  249-253. 


Fig.  249. 


Fig.  250. 


Fig.  251. 


Fig.  252. 


253- 


To  make  models  of  these  solids,  draw  on  cardboard  the  diagrams 
shown  in  these  figures,  cut  them  out  of  the  cardboard,  and  then 
cut  half  through  the  edges  of  the  adjacent  polygons.  The  figures 
will  then  readily  fold  up  into  the  shape  of  the  solids,  and  can  be 
retained  in  shape  by  means  of  glue  or  paste. 


Literally,  -unroll. 


§  244-] 


CHAPTER   XII.  —  GEOMETRICAL    BODIES. 


267 


Exercises.  —  1.    Give  examples  of  polyhedrons. 

2.  How  many  edges  have  each  of  the  regular  polyhedrons? 

3.  How  many  corners  have  each  of  the  regular  polyhedrons? 

4.  Verify  Euler's  theorem  for  each  of  the  regular  polyhedrons? 

5.  Construct  models  of  the  regular  polyhedrons. 

NOTE. —  In  making  the  model  of  the  regular  dodecahedron,  the  diagonals,  shown 
dotted  in  Fig.  253,  may  be  used  to  advantage. 

II.  — The  Prism,  Cylinder,  Pyramid,  and  Cone. 

§  244.  If  the  polygon  ABODE  (Fig.  254)  move  along  the 
line  AFt  keeping  parallel 
to  its  first  position,  its  sides 
will  describe  parallelograms, 
and  the  polygon  will  gener- 
ate a  solid  called  a  Prism. 

Definition.  —  A  PRISM 
is  a  polyhedron  bounded  by 
parallelograms  and  by  two 


Fig.  254. 


equal  and  parallel  polygons. 

The    parallelograms    are 
called  the  lateral  faces,  their  lines  of  intersection  the  lateral  edges, 
and  the  two  polygons  the  bases  of  the  prism. 

The  height  of  the  prism  is  the  distance  (HL)  between  the  bases. 

A  prism  is  called  triangular,  quadrangular,  etc.,  according  as 
the  base  is  a  triangle,  quadrilateral,  etc. 

A  prism  is  oblique  if  the  lateral  edges  are  inclined  to  the  bases 
(Fig.  255);  right,  if  they  are  normal  to  the  bases  (Fig.  256);  and 
regular,  if  it  is  a  right  prism  with  a  regular  polygon  for  its  base. 

A  parallelepiped  is  a  prism  whose  bases  are  parallelograms.  It 
is,  therefore,  a  solid  bounded  by  six  parallelograms. 


Fig.    255. 


Fig.  256.  Fig.  257.  Fig.  258. 


Fig.  259. 


268 


GEOMETRY    FOR    BEGINNERS. 


[§  MS- 


What  is  an  oblique  parallelepiped  (Fig.  257)  ?  A  right  parallele- 
piped (Fig.  258}  ? 

A  rectangular  parallelepiped  is  a  right  parallelepiped,  all  of 
whose  faces  are  rectangles. 

A  cube  is  a  right  parallelepiped,  all  of  whose  faces  are  squares 
(Fig.  259). 

Exercises.  —  1.  Can  you  give  an  example  of  a  prism?  What  kind  of 
a  prism  is  it  ? 

2.  In  Fig.  234,  point  out  lateral  faces,  lateral  edges,  the  bases,  the  height. 

3.  Name  a  prism  with  a  base  of  five  sides,  six  sides,  eight  sides. 

4.  What  are  the  lateral  faces  of  a  right  prism?     Why  is  a  lateral  edge 
equal  to  the  height  of  the  prism? 

5.  How  many  faces,  edges,  corners,  in  a  prism  with  a  base  of  three  sides, 
four  sides,  six  sides,  eight  sides,  twenty  sides?     Apply  Euler's  formula  to  each 
one  of  these  cases. 

6.  Draw  the  following :  — 

(*'.)     Oblique  triangular  prism.  (z>.) 

(z'z.)    Right  pentagonal  prism.  (&*•) 

(z'z'z.)  Right  hexagonal  prism.  (z/z'z.) 

(zV.)   Regular  hexagonal  prism.          (viii.} 


Oblique  parallelepiped. 
Right  parallelepiped. 
Rectangular  parallelepiped. 
Cube. 


245.  If  a  circle  (Fig.  260)  move  in  the  direction  normal  to 
its  own  plane,  keeping  always  parallel  to 
its  first  position,  it  generates  a  solid  called 
the  right  circular  Cylinder.  The  same 
solid  is  also  generated  by  a  rectangle 
ABCD  (Fig.  257),  which  revolves  about 
one  side  CD. 

Definition. — A  right  circular  CYL- 
INDER is  the  solid  produced  by  the  revolu- 
tion of  a  rectangle  about  one  of  its  sides. 

NOTE. —  In  future,  by  the  word  "cylinder"  the  right  circular  cylinder  will  be 


Fig.  260. 


meant. 


A  cylinder  is  bounded  by  a  curved  surface  called  the  convex 
surface,  and  by  two  equal  and  parallel  circles  called  the  bases. 
The  line  joining  the  centres  of  the  bases  is  called  the  axis ;  its 
length  is  equal  to  the  height  of  the  cylinder. 


§  246.] 


CHAPTER   XII. GEOMETRICAL    BODIES. 


269 


§  246.  If  the  polygon  ABODE  (Fig.  261}  move  along  the 
line  AO,  keeping  parallel 
to  its  first  position,  but 
diminishing  in  size  at  a 
uniform  rate,  without  al- 
tering •  in  shape,  until  it  is 
reduced  to  a  point  O,  its 
sides  will  describe  trian- 
gles, and  the  polygon  will 
generate  a  solid  called  a 


Pyramid. 


Fig.  261. 


Definition. — A  PYRAMID  is  a  polyhedron  bounded  by  triangles 
that  have  a  common  vertex,  and  by  a  polygon  whose  sides  are  the 
bases  of  the  triangles. 

The  triangles  are  called  the  lateral  faces  ;  their  lines  of  intersec- 
tion, the  lateral  edges ;  their  common  vertex,  the  vertex  of  the 
pyramid ;  and  the  polygon,  the  base  of  the  pyramid. 

The  heightQi the  pyramid  is  the  distance  (OF)  from  the  vertex 
to  the  base. 

A  pyramid  is  triangular,  quadrangular,  etc.,  ac- 
cording as  its  base  is  a  triangle,  quadrilateral,  etc. 

A  regular  pyramid  is  a  pyramid  which  has  a 
regular  polygon  for  its  base,  and  in  which  the 
perpendicular  from  the  vertex  passes  through  the 
centre  of  the  base  (Fig.  262).  The  distance  OM 
from  the  vertex  to  a  side  of  the  base  is  called 
the  slant  height. 


Fig.  262. 


Exercises.  —  1.    In   Fig.  261.  point  out  lateral  faces,  lateral  edges,  the 
base,  the  vertex,  the  height. 

2.  Name  a  pyramid  with  a  base  of  five  sides  ;   6  sides  ;   8  sides. 

3.  What  kind  of  a  pyramid  is  the  tetrahedron  (Fig.  244)  ? 

4.  What  kind  of  triangles  are  the  faces  of  a  regular  pyramid  ? 

5.  Draw  (*'.)  a  triangular  pyramid  ;    (it.}  a  regular  hexagonal  pyramid. 


270  GEOMETRY    FOR   BEGINNERS.  [§   247. 

§  247.    If  a  circle  (Fig.  263)  move  in  a  direction  normal  to  its 

own  plane,  keeping  parallel  to  its 
first  position,  but  diminishing  in 
size  at  a  uniform  rate  till  it  is 
reduced  to  a  point  O,  it  will  gener- 
ate a  solid  called  a  right  circular 
Cone. 

The  same  solid  is  also  generated 
Fig  263  by  a  right   triangle  AOB,  which 

revolves  about  the  leg  OB. 

Definition.  —  A  right  circular  CONE  is  the  solid  produced  by 
the  revolution  of  a  right  triangle  about  one  of  its  legs. 

NOTE.  —  In  future,  by  the  word  "  cone  "  a  right  circular  cone  will  be  meant. 

A  cone  is  bounded  by  a  curved  surface  called  its  convex  surface, 
and  by  a  circle  called  its  base.  The  point  O  is  the  vertex  of  the 
cone,  and  the  line  OB  joining  the  vertex  to  the  centre  of  the  base 
is  the  axis ;  its  length  is  equal  to  the  height  of  the  cone. 

The  length  of  the  line  AO  (Fig.  263}  is  called  the  slant  height. 

§  248.  From  the  mode  of  generation  of  the  four  bodies  now 
described,  —  Prism,  Cylinder,  Pyramid,  Cone,  —  it  follows  that  a 
section  made  by  a  plane  parallel  to  the  base  is,  — 


\ ': 


Fig.  264.  Fig.  265.  ~  Fig.  266. 


In  a  prism,  a  polygon  equal  to  the  base. 

In  a  cylinder,  a  circle  equal  to  the  base  {Fig.  264). 

In  a  pyramid,  a  polygon  similar  to  the  base  (Fig.  271} . 

In  a  cone,  a  circle  smaller  than  the  base  (Fig.  272). 

The  section  of  a  cylinder  made  by  a  plane  inclined  to  the  base 


§   249-]  CHAPTER   XII. — GEOMETRICAL   BODIES.  271 

is  an  ellipse  (Fig-  265} .  Such  sections  are  made  when  two  cylin- 
ders intersect  each  other,  as  in  the  case  of  two  pipes  meeting  at 
right  angles  (Fig.  266) . 

In  a  cone,  a  section  inclined  to  the  base  cuts  the  surface  in 
an  ellipse  (Fig.  268} ,  if  it  does  not  meet  the  base ;  in  a  parabola 
(Fig.  269),  if  it  meets  the  base  and  is  parallel  to  the  side  of  the 
cone  ;  in  a  hyperbola  (Fig.  270) ,  if  it  meets  the  base  and  is  not 
parallel  to  the  side  of  the  cone. 


Fig.  267.  Fig.  268, 

The  four  sections  shown  in  Figs.  267-270  —  namely,  circle, 
ellipse,  parabola,  hyperbola  —  are  sometimes  called  the  conic 
sections. 

NOTE.  —  These  sections  may  be  shown  to  the  eye  by  holding  a  cone  partly 
under  water,  at  first  with  the  axis  vertical,  then  with  the  axis  more  and  more 
inclined.  If  the  water  is  colored  red,  the  effect  is  better  seen. 

Exercises.  — What  is  the  section  made  by  a  plane  passing  through,  ^ — 

1.  Two  lateral  edges  of  a  prism  ?  3.    The  axis  of  a  cylinder  ? 

2.  Two  lateral  edges  of  a  pyramid  ?        4.   The  axis  of  a  cone  ? 

§  249.  A  plane  parallel  to  the  base  of  a  pyramid,  or  of  a  cone, 
divides  the  body  into  a  smaller  pyramid  or  cone,  and  a  body  called 
the  frustum  of  the  pyramid  or  cone  (Figs.  271  and  272) .  The 
base  and  the  section  parallel  to  the  base  are  called  the  lower  and 
upper  bases  of  the  frustum ;  their  distance  from  each  other  is  the 
height  of  the  frustum. 

If  we  know  the  height  PQ  (Fig.  271}  of  a  frustum  of  a  pyramid, 
and  two  homologous  sides  AB  and  EF  of  the  bases,  we  can  find 
the  height  OP  of  the  entire  pyramid,  as  follows  :  — 

Through  OP  and  the  edge  OA  pass  a  plane  intersecting  the 


272 


GEOMETRY   FOR    BEGINNERS. 


[§  250. 


bases  of  the  frustum  in  the  parallel  lines  APand  EQ  (why  paral- 
lel?), and  draw  E  M  II  OB.     By  §  159,  — 


=_       d   OE  = 
'  ' 


PQ      EA'  EA      AM 

Therefore,  — 

OQ      EF 
=  ™ 


And 


PQx  EF     PQxEF 
=        AUT 
OP=OQ+QP. 


Fig.  271. 


Fig.  272. 


Exercises.  —  1.    What  are  the  faces  of  the  frustum  of  a  pyramid  ? 

2.  If  (Fig.  271}  PQ  =  Sm,  A£=4m,  EF=  2m,  find  OQ  and  OP. 

3.  Given  the  height  PQ  (Fig.  272}  of  the  frustum  of  a  cone,  and  the  radii 
of  the  bases  ;   find  the  height  of  the  cone  cut  off,  and  also  the  height  of  the 
entire  cone. 

Hint.  — Draw  BC  \\  OP,  and  use  the  similar  triangles  OBQ  andA£C. 

4.  If  (Fig.  272}  PQ=4m,  AP  =  3m,  BQ_  =  im,  find  OQ  and  OP. 

§  250.  The  surfaces  of  the  prism,  the  pyramid,  the  cylinder,  and 
the  cone  are  all  developable. 

Fig.  273  shows  the  development  of  a  regular  hexagonal  prism ; 
Fig.  2J4  that  of  a  regular  pyramid  whose  base  is  a  square. 

The  auxiliary  lines  in  the  two  figures  serve  to  illustrate  how  they 
are  to  be  constructed.  Fig.  273  is  the  development  of  a  regular 
hexagonal  pyramid.  How  is  it  constructed  ? 


250.] 


CHAPTER   XII. GEOMETRICAL    BODIES. 


273 


The  development  of  the  convex  surface  of  a  cylinder  is  a  rect- 
angle ;  for  we  can  roll  up  a  rectangular  sheet  of  paper,  for  exam- 
ple, into  the  shape  of  a  cylindrical  surface  ;  and  conversely,  a  cylin- 
drical surface  can  be  unrolled  into  a  rectangle  (Fig.  276). 


Fig.  274. 


275- 


Fig.  276. 


Fig.  277. 


Fig.  278. 


The  development  of  the  convex  surface  of  a  cone  is  a  circular 
sector  whose  arc  is  the  circumference  of  the  base  of  the  cone,  and 
whose  radius  is  the  slant  height  of  the  cone  (Fig.  278). 

Fig.  276  shows  the  development  of  the  entire  surface  of  a  cylin- 
der ;  Fig.  277,  that  of  the  entire  surface  of  a  cone ;  Fig.  277,  that 
of  the  entire  surface  of  the  frustum  of  a  cone. 

Exercises.  —  Make  models  of  the  following  solids :  — 


1.  A  regular  triangular  prism. 

2.  A  regular  hexagonal  prism. 

3.  A  right  pi-ism,  not  regular. 
4:,  A  rectangular  parallelepiped. 
5*  A  regular  hexagonal  pyramid. 


6.  A  cube. 

7»  A  cylinder. 

8.  A  cone. 

9.  A   frustum  of  a  pyramid. 
10.  A  frustum  of  a  cone. 


274 


GEOMETRY    FOR    BEGINNERS. 


[§   25L 


Fig.  279. 


III.  —  The  Sphere. 

§  251.    Definition.  —  A  SPHERE  is  the  solid  produced  by  the 
revolution  of  a  semicircle  (or  circle)  about  the  diameter  (Fig.  279). 

NOTE.  —  Spin  a  piece  of  money  rapidly  on  the  table,  and  it  will  present  the 
appearance  of  a  sphere. 

The  semi-circumference  PMQ  describes  the  surface  of  the 
sphere  ;  and  it  is  clear  that  this  surface 
must  be  everywhere  equally  distant 
from  O,  the  centre  of  the  semicircle 
This  point  O  is  called  the  centre  of  th< 
sphere. 

Therefore,  a  sphere  may  also  be  de- 
fined as  a  solid  bounded  by  a  surface, 
which    is   everywhere   equally   distanl 
from  a  point  called  the  centre. 

What  is  a  radius  of  a  sphere  ?  a  diameter  ? 
An  orange  may  be  taken  as  a  good  example  of  a  sphere.  If  we 
cut  an  orange  into  a  series  of  parallel  slices, 
the  planes  of  division  between  the  slices 
are  circles  differing  in  size,  the  largest  being 
that  which  passes  through  the  centre  of  the 
orange. 

Every  section  of  a  sphere  made  by  a  plane 
is  a  circle. 

If  the  section  passes  through  the  centre 
of  the   sphere,  it  is   called  a  great  circle ; 
in  all  other  cases,  it  is  called  a  small  circle. 
The  equal  parts  into  which  a  great  circle  divides  the  sphere  are 
called  Hemispheres. 

In  Fig.  2 So,  the  sections  PAQB  and  AGB H are  great  circles ; 
the  sections  CEDF2cs\&  MS  NT  me  small  circles. 

All  great  circles  of  a  sphere  are  equal;  they  have  the  same  centre, 
the  same  radius,  the  same  diameter,  as  the  sphere  itself. 


§  251.]  CHAPTER   XII.  —  GEOMETRICAL   BODIES.  275 

All  great  circles  divide  each  other  into  two  equal  parts ;  every 
great  circle  divides  the  sphere  and  its  surface  into  two  equal  parts. 

Small  circles  are  the  smaller  the  further  they  are  from  the  centre 
of  the  sphere. 

Every  small  circle  divides  the  sphere  and  its  surface  into  two 
unequal  parts. 

The  centre  of  any  circle  of  the  sphere  must  lie  in  the  diameter 
of  the  sphere  which  is  normal  to  the  plane  of  the  circle ;  the  two 
points  where  this  diameter  meets  the  surface  of  the  sphere  are  called 
the  Poles  of  the  circle.  In  Fig.  280,  the  points  P  and  Q  are  the 
poles  of  what  circles  ? 

On  the  earth  (or  terrestrial  sphere) ,  parallels  of  'latitude  are  small 
circles,  the  equator  is  a  great  circle,  and  the  north  and  south  poles 
are  the  poles  of  all  these  circles. 

All  points  in  the  circumference  of  a  circle  (great  or  small)  are 
equidistant  from  either  one  of  its  poles,  whether  this  distance  be 
measured  in  a  straight  line  or  along  the  arc  of  a  great  circle  pass- 
ing through  the  point  and  the  pole.  Since  the  latter  way  of  meas- 
uring distances  is  much  more  convenient  on  a  sphere,  it  is  always 
employed.  Thus,  in  Fig.  280,  the  distance  from  the  point  C  to 
the  pole  Pis  the  length  of  the  arc  PC. 

Of  all  lines  that  can  be  drawn  on  the  surface  of  a  sphere  from 
one  point  to  another,  the  arc  of  the  great  circle  passing  through 
the  two  points  is  the  shortest.  In  this  respect,  the  arc  of  a  great 
circle  is  to  the  surface  of  a  sphere  what  the  straight  line  is  to  the 
plane. 

By  means  of  arcs  of  great  circles  (never  small  circles) ,  spherical 
triangles  and  spherical  polygons  may  be  drawn  on  the  surface  of  a 
sphere. 

The  portion  of  the  surface  of  a  sphere  comprised  between  the 
circumference  of  two  parallel  circles  is  called  a  Zone.1  The  dis- 


1  From  a  Greek  word,  meaning  belt  or  girdle. 


276  GEOMETRY   FOR   BEGINNERS.  [§   252. 

tance  between  the  planes  of  the  circles  is  the  height  of  the  zone, 
and  the  circumferences  of  the  circles  are  its  bases.  If  one  of 
the  planes  touches  the  sphere  without  cutting  it  (or  is  a  tangent 
plane),  the  zone  is  called  a  zone  of  one  base.  Point  out  zones  in 
Fig.  280. 

Exercises.  —  1.    Give  examples  of  spheres  (globes,  balls). 

2.  Draw  a  figure  of  a  sphere  with  lines  in  it  to  illustrate  the  definitions  of 
this  section. 

3.  How  will  a  sphere  be  divided  by  three  great  circles  mutually  perpen- 
dicular to  each  other  ? 

4.  In  what  zones  is  the  surface  of  the  earth  divided  ?     Which  are  zones 
of  one  base  ? 

§  252.  An  orange,  after  the  peel  is  removed,  is  easily  divided 
into  wedge-shaped  slices,  separated  by  natural  planes  of  division. 
These  planes  are  planes  of  great  circles  intersecting  in  a  common 
diameter.  In  the  fruit,  the  number  of  the  planes  is  limited  ;  but, 
in  any  sphere,  we  may  imagine  as  many  such  planes  as  we  please, 
all  passed  through  a  common  diameter,  and  each  cutting  the  sur- 
face of  the  sphere  in  the  circumference  of  a  great  circle. 

Fig.  281  shows  several  such  planes  passing  through  the  diameter 
PQ.  They  intersect  the  surface  in  the 
circumferences,  the  front  parts  of  which 
are  the  semi-circumferences  PA  Q,  PBQ, 
etc. 

These  semi-circumferences  are  called 
Meridians,  and  the  portion  of  the  sur- 
face of  the  sphere  comprised  between  two 
meridians  is  called  a  Lune.  Point  out 
lunes  in  Fig.  281. 

Terrestrial  Meridians  are  the  circum- 
ferences of  great  circles  on  the  earth's  surface  made  by  planes 
passing  through  the  earth's  axis. 

The  surface  of  a  sphere  is  not  developable,  but  a  close  approxi- 


CHAPTER   XII. GEOMETRICAL    BODIES. 


277 


mation  to  it  can  be  constructed  by  drawing  a  series  of  equal  lunes 
in  one  plane,  as  follows  :  — 

Divide  a  line  AB  {Fig.  282),  which  must  be  made  equal  to  3^ 
times  the  diameter  of  the  proposed  sphere,  into  twelve  equal  parts, 
and  add  nine  of  these  parts  to  each  end  of  the  line.  Then,  with  a 


U 


A 


V 


A 


A 


\l 


i  ii  in 


t-M- 


Fig.  282. 


radius  equal  to  ten  of  these  parts,  describe  arcs  from  the  points 
marked  i,  2,  3,  etc. ;  and,  likewise,  arcs  from  the  points  marked 
I.,  II.,  III.,  etc.,  as  shown  in  the  figure.  These  arcs  will  inclose 
lunes  which,  cut  out  and  properly  folded  up,  will  very  nearly  form 
a  complete  spherical  surface. 

It  is  by  a  process  of  this  sort  that  the  surfaces  of  balloons,  base- 
balls, globes,  etc.,  are  made. 

Exercises.  —  1.    Describe    how   a   hemisphere   would   be    generated   by 
motion. 

2.    Make  a  model  of  a  sphere. 


278  GEOMETRY    FOR   BEGINNERS. 


REVIEW   OF   CHAPTER  XII. 

I.  —Polyhedrons. 

§  242.    Definition  of  the  Polyhedron.    Specific  names  of  polyhedrons.    Euler's 

theorem. 
§  243.   The  five  regular  polyhedrons.     Development  of  their  surfaces.     How 

models  of  them  are  made. 

II. —The  Prism,  Cylinder,  Pyramid,  and  Cone. 

§  244.    Definition  of  the  Prism.     Different  kinds  of  prisms.     The  parallele- 
piped. 

§  245.    Definition  of  the  Cylinder. 

§  246.   Definition  of  the  Pyramid.      Different  kinds  of  pyramids. 
§  247.   Definition  of  the  Cone. 

§  248.    Sections  of  the  prism,  cylinder,  pyramid,  and  cone. 
§  249.    Frustum  of  a  pyramid  or  a  cone. 

§  250.   Development  of  the  surfaces  of  these  four  solids.     How  models  of 
them  are  made. 

III. —The  Sphere. 

§  251.  The  Sphere  defined.     Section  of  a  sphere  made  by  a  plane.     Great 

and  small  circles.     Spherical  triangles  and  polygons.     Zones. 
§  252.   Meridian  lines.     Lunes.     How  a  model  of  a  sphere  may  be  made. 


§   253-]     CHAPTER  XIII.  —  SURFACES  AND  VOLUMES  OF  BODIES.  279 


CHAPTER  XIII. 
SURFACES    AND    VOLUMES    OP    BODIES. 

CONTENTS.  —  I.  Surfaces  of  Bodies  (§§  253-258).     II.  Volumes  of  Bodies  ($$259- 
274).     III.  Exercises  and  Applications  (§§  275-279). 

J.  —  Surfaces  of  Bodies. 

§  253.   THE  PRISM.     The  lateral  surface  consists  of  parallelo- 
grams ;  the  bases  are  polygons.     Therefore,  — 

I.  Lateral  surface  =  sum  of  the  areas  of  the  parallelograms  of 
which  it  is  composed. 

II.  Total  surface  =  lateral  surface  -\-  areas  of  the  bases. 

Exercises.  —  Find  the  total  surface  of  a  cube  if,  — 

1.  The  edge  =  8m.  3.   The  edge  =  4ft. 

2.  The  edge  =  3.5m.  4.   The  edge  =  7fin. 

5.  If  the  edge  of  a  cube  contain  a  unit  of  length,  how  many  units  of  area 
are  there  in  the  entire  surface  ?  Ans.   6  a2. 

6.  The  total  surface  of  a  cube  —  6ooim.     Find  the  length  of  the  edge. 

7 .  If  the  total  surface  of  a  cube  contains  S  units  of  area,  how  many  units 
of  length  are  there  in  one  edge  ?  Ans.    -\J  S  -+  6. 

8.  Find  the  total  surface  of  a  rectangular  parallelepiped  if  the  length  = 
15™,  the  breadth  =  9m,  the  height  —  6m. 

9.  Find  the  total  surface  of  a  rectangular  parallelepiped  if  the  length  =  a, 
breadth  =  b,  height  —  c.  Ans.    2  ab  +  2  ac  -f  2  be. 

10.  How  many  bricks,  8in  by  3!™,  are  required  to  line  the  bottom  and  sides 
of  a  reservoir  60^  long,  24**  wide,   12^-  deep  ? 

Find  the  total  surface  of  the  following  regular  prisms :  — 

11.  Triangular  prism,  height  =  76cm,  side  of  base  =  40°™. 

12.  Quadrangular  prism,  height  =  6ft,  side  of  base  =  27in. 

13.  Hexagonal  prism,  height  =  2.46m,  side  of  base  —  84cm. 

Hints.  —  Find  less  radius  of  base  by  use  of  $  202,  Case  2,  and  $  142,  and  area 
of  base  by  $  133. 


280  GEOMETRY    FOR    BEGINNERS.  [§   254. 

14,  How  would  Exercise  13  be  solved  if  the  less  (or  the  greater)  radius 
of  the  base  were  given  instead  of  a  side  of  the  base  ? 

15,  The  total  surface  of  a  regular  hexagonal  prism  =  822<im,  a  side  of  the 
base  —  iom.     Find  the  height  (see  Table,  p.  221). 

16,  In  a  regular  hexagonal  prism,  the  height  =  /£,  a  side  of  the  base  =  a. 
Find  the  total  surface.  Ans.    6a/i+  3«2V3- 

§  254.  THE  CYLINDER.  The  development  of  the  convex  sur- 
face is  a  rectangle  {Fig.  276}  whose  base  is  equal  to  the  circum- 
ference of  the  base  of  the  cylinder,  and  whose  altitude  is  equal  to 
the  height  of  the  cylinder.  The  bases  of  the  cylinder  are  equal 
circles.  Therefore  (§  128),- 

I.  Convex  surface  =  circumference  of  base  X  height  of  cylinder. 

II.  Total  surface  —  convex  surface  +  twice  the  area  of  either 
base. 

Exercises.  —  Find  the  convex  surface,  the  areas  of  the  bases,  and  the  total 
surface  of  the  following  cylinders  :  — 

1.  Height  =  3m,  radius  of  base  =  im. 

2.  Height  =  96cm,  diameter  of  base  =  6ocm. 

3.  Height  —  4.5m,  circumference  of  base  =  3.64m. 

4.  Height  =  34m,  area  of  base  =  i6.74<im. 

5.  Height  =  9^ft,  radius  of  base  —  6ft. 

6.  Height  =  h,  radius  of  base  =  r. 

7.  Height  =  h,  diameter  of  base  —  d. 

8.  How  many  sq.  yds.  of  sheet  tin  are  required  to  cover  the  convex  surface 
of  a  circular  tower  40**  high,  the  diameter  of  the  base  being  8ft? 

9.  The  length  of  a  hollow  tube  =  3m,  the  outer  diameter  =  36°™,  the  thick- 
ness of  the  metal  =  2cm.     Find  the  area  of  the  inner  convex  surface. 

§  255.  THE  PYRAMID.  The  lateral  surface  consists  of  triangles  ; 
the  base  is  a  polygon.  Therefore,  — 

I.  Lateral  surface  =  sum  of  the  areas  of  the  triangles  of  which 
it  is  composed. 

II.  Total  surface  =  lateral  surface  -f  area  of  the  base. 
Exercises.  —  Find  the  total  surface  of  the  following  regular  pyramids :  — 
1,   Triangular  pyramid,  slant  height  =  68cm,  side  of  base  —  42°™. 


§256.]     CHAPTER  XIII. SURFACES  AND  VOLUMES  OF  BODIES.  281 

2.  Square  pyramid,  slant  height  =  1 5™,  side  of  base  =  3m. 

3.  Octagonal  pyramid,  slant  height  =  7.o6in,  side  of  base  =  1. 44™. 

4.  How  can  the  above  exercises  be  solved  if  the  less  (or  greater)  radius 
of  the  base  is  given  in  place  of  a  side  ?     (See  Table,  p.  221.) 

5.  The  height  of  a  regular  triangular  pyramid  =  8m,  and  the  side  of  the 
base  =  2m.     Find  the  total  surface. 

Hints.  —  Find  less  radius  of  base  by  means  of  the  Table,  p.  221,  and  then  the 
slant  height  by  means  of  §  142. 

6.  Height  of  a  regular  hexagonal  pyramid  =  i8m,  side  of  the  base  =  i-5m. 
Find  the  entire  surface. 

7.  The  vertex  of  a  pyramid  with  a  rectangular  base  is  vertically  above  the 
middle  point  of  the  base.     Find  the  total  surface  if  the  height  =  6m,  and  the 
sides  of  the  base  are  2.4'"  and  i.6m. 

8.  The  height  of  the  frustum  of  a  regular  pyramid  with  a  square  base  = 
l6m,  and  the  sides  of  the  bases  are  4™  and  2m.     Find  the  entire  surface. 

Hints.  — 'The.  lateral  surface  consists  of  four  equal  trapezoids.  Find  their  com- 
mon altitude  by  means  of  §  142. 

§  256.  THE  CONE.  The  developed  convex  surface  is  a  circular 
sector  {Fig.  277}  whose  arc  is  equal  to  the  circumference  of  the 
cone,  and  whose  radius  is  equal  to  the  slant  height.  The  base  of 
the  cone  is  a  circle.  Therefore  (§  216),  — 

I.  Convex   surface  =  circumference    of  base  X  half  the   slant 
height. 

II.  Total  surface  =  convex  surface  -f-  area  of  the  base. 

Exercises.  —  Find  the  convex  surface,  the  area  of  the  base,  and  the  entire 
surface  of  the  following  cones  :  — 

1.  Slant  height  =  5™,  radius  of  base  =  2m. 

2.  Slant  height  =  60^,  diameter  of  base  =  I2ft. 

3.  Slant  height  =  I2.62m,  circumference  of  base  —  2.27m. 

4.  Slant  height  —  k,  radius  of  base  =  r. 

5.  Slant  height  =  k,  diameter  of  base  =  d. 

6.  Height  =  I2m,  radius  of  base  =  5™  (see  §  142). 

7.  Height  =  /i,  radius  of  base  =  r. 

8.  How  many  square  yards  of  sail-cloth  are  required  to  make  a  conical 
tent  6oft  high,  if  the  diameter  of  the  base  is  40**. 

9.  Two  cones  have  equal  bases,  but  one  is  three  times  as  high  as  the  other. 
Compare  their  convex  surfaces. 


282 


GEOMETRY   FOR   BEGINNERS. 


[§  257- 


§  257.  FRUSTUM  OF  A  CONE.  The  development  of  this  surface 
is  a  part  of  a  circular  ring  (Fig.  283)  whose  arcs  A B  and  CD  are 

the    circumferences   of  the 

//Yx  bases   of  the   frustum,  and 

/    /    \    \  whose     distance     apart     is 

equal  to  the  slant  height 
of  the  frustum.  We  shall 
first  find  the  area  by  the 
method  of  reasoning  used 
in  §  212  (a  method  which 
we  shall  have  occasion  to 
use  again  hereafter);  and 
shall  then  transform  the 
value  of  the  area  thus  obtained  into  two  other  useful  forms. 

Construct  a  series  of  trapezoids,  such  as  the  three  shown  in  Fig. 
283,  by  drawing  radii  and  parallel  pairs  of  tangents  (why  are  they 
parallel  ?) .  The  common  altitude  of  these  trapezoids  =  distance 
between  the  arcs  —  slant  height  of  frustum.  By  increasing  the 
number  of  such  trapezoids  we  can  plainly  make  the  sum  of  their 
areas  approach  the  area  of  the  frustum  as  nearly  as  we  please, 
although  it  will  never  quite  reach  it.  That  is  to  say,  — 

Surface  of  the  frustum  =  limit  of  the  sum  of  the  trapezoids. 
By  §  131  and  by  addition,  the  sum  of  the  trapezoids  =  £  sum 
of  their  parallel  sides  X  slant  height.     As  the  number  of  trapezoids 
is  increased,  the  sum  of  their  parallel  sides  approaches,  but  never 
quite  reaches,  the  sum  of  the  arcs  AB  and  CD.     Therefore,  — 

{±(AB  +  CD)  X  slant  height 
Limit  of  sum  of  the  trapezoids  =  -H 

(.  of  frustum. 

Therefore  (Axiom  I.), — 

I.  Convex  surface  of  frustum  =  £(AJ3  X  CD)  X  slant  height 
=  £(sum  of  circumferences  of  bases)  X  slant  height. 

The  result  can  be  put  into  another  form,  as  follows  :  Let  a  frus- 
tum be  generated  by  the  revolution  of  the  trapezoid  AX  BY  (Fig. 


§257-]     CHAPTER  Xlil.  —  SURFACES  AND  VOLUMES  OF  BODIES.  283 

284)  about  the  side  XY.     Through  M,  the  middle  point  of  AS, 
draw  MN\\  AX.     MN  is  the  radius  of  a  circu- 
lar  section   of  the  frustum  midway  between  the 
bases.     Draw  BP  II  XY,  and  intersecting  MN 


(why?).     Put  BY=  a,  MQ  =  b  ;  then  MN  = 
a+b,AX=a  +  2b.    By  §209,- 

Flg.  284. 

Circumference  of  upper  base       =  2  TT  a. 
"  lower     " 


11  middle  section  =  2ira  +  2-n-b. 
Whence,    £  (sum   of  circumferences    of  bases)  =  4 


=  2Tra  -f-  2Trb  =  circumference  of  middle  section. 

Therefore,  — 

II.  Convex  surface  of  frustum  =  circumference  of  middle  section 
X  slant  height. 

Draw  MR\.AB\  AMNR  ~  AA£P  (why?);  therefore, 
MN:  MR  =  BPor  its  equalXY:  AB;  whence,— 

MN  x  AB  =  MR  x  XY. 

If  we  multiply  both  sides  of  this  equation  by  2  TT,  we  have 
AB=  2-rrMR  x  XY. 


The  first  side  of  this  equation  is  the  value  of  the  convex  surface 
given  in  II.  ;  the  second  side  is  the  product  of  the  circumference 
whose  radius  is  MR  and  the  height  of  the  frustum.  Therefore,  — 

III.  Convex  surface  of  frustum  =  circumference  whose  radius 
is  the  length  of  the  perpendicular  erected  at  the  middle  point  of  the 
side  of  the  frustum  and  terminated  by  the  axis  X  height  of  frustum. 

Exercises.  —  1.  Prove  that  Theorems  II.  and  III.  hold  true  of  the  convex 
surface  of  an  entire  cone. 

Hint.  —  In  the  proof  of  II.,  make  a  ~  o,  because  there  is  no  upper  base. 

2.  Prove  that  Theorem  III.  holds  true  of  the  convex  surface  of  a  cylinder. 

3.  How  is  the  total  surface  of  a  frustum  found  ? 


284 


GEOMETRY    FOR    BEGINNERS. 


[§  258. 


Find  the  convex  surface,  and  also  the  total  surface,  of  the  following  frus- 
tums of  cones  :  — 

4.  Slant  height  =  I  im,  radii  of  bases  5™  and  3m. 

5.  Slant  height  =  4ft,  diameter  of  bases  i8in  and  ioin. 

6.  Slant  height  =  k,  radii  of  bases,  a  and  b. 

7.  Slant  height  =  k,  radius  of  middle  section  —  r. 

8.  The  height  of  a  frustum  =  I2m,  the  radii  of  the  bases  are  ym  and  2*. 
Find  the  convex  surface. 

Hint.  —  It   is   evident,   from   Fig.   284  and   from    §    142,   that   slant   height  = 


9,   The  height  of  a  frustum  —  h,  the  radii  of  the  bases  are  a  and  b.     Find 
the  convex  surface. 


258.  THE  SPHERE.  Inscribe  in  the  semicircle  X BY  (Fig. 
with  a  radius  OX—  r,  half  of  a  regular  polygon  with  an 
even  number  of  sides  (here  four).  Draw  the 
less  radii  OM,  ON,  OP,  OQ,  and  from  the 
vertices  A,  B,  C,  let  fall  the  perpendiculars 
AD,  BO,  CE. 

If  the  semi-circumference  and  the  sides  of 
the  polygon  revolve  about  X  Y,  the  former  will 
describe  the  surface  of  a  sphere,  and  the  sides 
of  the  polygon  will  describe  the  convex  surfaces 
of  cones  or  frustums  of  cones.  Which  two 
sides  will  describe  the  former? 

By  §  257,  III.  (including  Exercise  i),— 

Surface  described  by  side  XA  =  z-n-OM  x  DX. 

X  OD. 
X  EO. 
"  "  "       CY=  2irOQ   x   YE. 

Adding  these  equations,  and  remembering  that  OM=ON  — 
OP=OQ,we  have, — 

Total  surface  described  by  sides  of  polygon  =  2irOM  X  XY. 
As  the  number  of  sides  is  increased,  the  total  surface  described 


Fig.  285. 


§258.]     CHAPTER  XIII.  —  SURFACES  AND  VOLUMES  OF  BODIES.  285 

by  the  sides  approaches  the   surface   of  the   sphere   as  its  limit. 
That  is,  - 

( limit  of  surface  described  by  sides  of  the 
Surface  of  sphere  =  < 

(      polygon. 

As  the  number  of  sides  is  increased,  the  less  radius  O  M  ap- 
proaches r  as  its  limit.  Therefore,  — 

Limit  of  surface  described  by  }  vv 

r    —    2  7T  r   X  -A   X . 

the  sides  of  the  polygon       J 

Therefore  (Axiom  I.), — 

Surface  of  the  sphere  =  2tcr  X  XY=  2-n-rX  2r  =  ^irr2. 

Or  (§§209,213),- 

I.  Surface  of  a  sphere  =  circumference  of  a  great  circle  X  the 
diameter  ^=four  times  the  area  of  a  great  circle. 

The  above  method  applied  to  a  part  of  the  semicircle,  —  as,  for 
instance,  the  arc  ABC,  which,  by  revolving,  describes  a  zone, — • 
gives  as  the  result,  — 

II.  Surface  of  a  zone  =  circumference  of  a  great  circle  X  height 
of  zone. 

Exercises.  —  Find  the  surfaces  of  the  following  spheres :  — 

1.  Radius  =  5™.  4,    Diameter  —  44™. 

2.  Radius  =  3|in.  5.    Diameter  =  I2ft. 

3.  Radius  =  r.  6.    Diameter  =  d. 

7.  Find  the  surface  of  the  earth,  taking  the  diameter  as  8000  miles. 

8.  If  the  surface  of  a  sphere  =  4001™,  find  the  radius. 

9.  Find  the  total  surface  of  a  hemisphere  if  the  radius  =  6m. 

10.  Find  the  total  surface  of  a  hemisphere  if  the  radius  =  r. 

11.  How  much  leather  will  it  take  to  cover  a  base-ball  if  the  diameter  = 
4p? 

12.  If  the  radius  of  a  sphere  =  8m,  find  the  area  of  a  zone  the  height  of 
which  =  2m. 

13.  Height  of  a  cylinder  =  diameter  of  its  base.     Compare  its  total  sur- 
face with  that  of  the  largest  sphere  that  can  be  inscribed  in  it. 

Ans.    Surface  of  the  sphere  =  f  surface  of  the  cylinder. 
NOTE.  —  This  curious  fact  was  discovered  by  Archimedes  more  than  200  B.C. 


286  GEOMETRY    FOR    BEGINNERS.  [§   259. 

II.—  Volumes  of  Bodies. 

§  259.  In  order  to  compare  the  sizes  or  magnitudes  of  bodies, 
we  select  a  body  of  known  size  as  the  common  term  of  comparison 
or  unit,  and  then  find  how  many  times  this  unit  is  contained  in 
other  bodies. 

Definition.  —  The  number  of  times  the  unit  is  contained  in  a 
body,  followed  by  the  name  of  the  unit,  is  called  the  VOLUME  of  the 
body. 

If,  for  instance,  the  cubic  inch  is  the  unit,  and  one  body  will 
contain  it  four  times  while  another  body  will  contain  it  twelve 
times,  the  volumes  of  the  bodies  are  four  cubic  inches  and  twelve 
cubic  inches ;  and  one  of  the  bodies  is  exactly  three  times  as  large 
as  the  other. 

The  most  convenient  units  of  volume  are  cubes  whose  edges  are 
equal  to  the  units  of  length. 

The  CUBIC  INCH  (cub.  in.),  the  CUBIC  FOOT  (cub.  ft.),  and  the 
CUBIC  YARD  (cub.  yd.),  are  such  units.  The  BUSHEL  and  the 
GALLON  are  not  such  units.  The  bushel  contains  2150.42  cub.  in., 
the  (United  States)  liquid  gallon  contains  231  cub.  in.,  and  the 
(United  States)  dry  gallon  contains  268.8  cub.  in. 

In  the  metric  system,  the  units  of  volume  are  always  cubes 
whose  edges  are  units  of  length.  The  three  most  used  are  the 
CUBIC  METER  (cbm.),  the  CUBIC  DECIMETER  (cdm.),  and  the 
CUBIC  CENTIMETER  (ccm.). 

The  cubic  decimeter  is  often  called  a  LITER  (1.).  A  HECTO- 
LITER (hi.)  =  100  liters. 

The  cubic  meter,  when  used  for  measuring  wood,  is  called  a 
STERE. 

NOTE.  —  i  cubic  foot      =  0.0283  cubic  meter.     I  cubic  meter  —  35.3156  cubic  feet. 
I  bushel  =  35.24  liters.  i  liter  =  0.02838  bushel, 

i  liquid  gallon  =  3.785  liters.  I  liter  —  0.264  gallon. 

Exercises.  —  1.  The  bushel  contains  4  pecks.  How  many  cubic  inches  in 
one  peck  ? 


§    260.]     CHAPTER  XIII. — SURFACES  AND  VOLUMES  OF  BODIES.  287 

2.  The  gallon  of  either  kind  is  divided  into  four  quarts,  the  quart  into  two 
pints,  the  pint  into  four  gills.     How  many  cubic  inches  in  the  quart,  pint,  and 
gill,  both  liquid  and  dry  ? 

3.  Reduce  6000  cubic  feet  to  cubic  meters.    6000  cubic  meters  to  cubic  feet. 

4.  If  I  pay  $1.20  per  bushel  for  wheat,  what  must  I  ask  per  hectoliter,  in 
order  to  make  a  profit  of  25  per  cent  ? 

5*    A  hectoliter  of  wine  will  fill  how  many  pint  bottles  ? 

§  260.  To  measure  the  volume  of  a  body  is  to  find  how  many 
times  it  will  contain  the  unit  of  volume. 

The  volume  of  a  liquid  body  may  be  found  directly  by  repeat- 
edly filling  a  vessel  which  is  known  to  hold  exactly  a  unit ;  as,  for 
example,  a  quart  pot  or  a  gallon  jug.  But  this  method  could  not 
be  applied  to  bodies  of  fixed  shape,  such  as  a  stick  of  timber  or  a 
cannon-ball. 

Geometry  teaches  how  volumes  are  measured  indirectly,  by 
showing  that  they  depend  upon  the  lengths  of  certain  lines  (for 
instance,  the  length,  breadth,  and  height  of  a  parallelepiped),  and 
can,  therefore,  be  found  by  performing  simple  operations  upon 
the  "numbers  which  express  the  lengths  of  these  lines  (compare 
§126). 

We  shall  deduce  a  general  rule  for  computing  the  volume  of  any 
prism  or  cylinder  by  treating  as  special  cases,  — 

(z.)     The  rectangular  parallelepiped  ; 

(«.)    The  right  parallelepiped  ; 

(/«.)  The  oblique  parallelepiped ; 

(iv.)  The  triangular  prism  ; 

(v.)     Any  prism ; 

(vi.)  The  cylinder ; 

and  another  general  rule  for  computing  the  volume  of  any  pyramid 
or  cone  by  treating  as  special  cases,  — 

(/.)     The  triangular  pyramid ; 

(«.)    Any  pyramid ; 

(in-)  The  cone. 


288 


GEOMETRY    FOR    BEGINNERS. 


[§   26l. 


Fig.  286. 


§  261.  THE  RECTANGULAR  PARALLELOPIPED.  In  the  rectangu- 
lar parallelepiped  AP  (Fig.  286)  let  the  edges  be  AB  =  5™,  AD 
=  3m,  AM=  7m.  Then  the  base  A  BCD  con- 
tains i5qm  (§  128);  and  upon  each  square  me- 
ter of  the  base  we  can  construct  a  pile  of  7 
cubic  meters,  such  as  DQ,  so  that  the  solid 
must  contain  exactly  7  X  15  =  105  cubic  me- 
ters. 

A  like  result  would  be  obtained  if  the  edges 
had  any  other  values.     If  there  were  fractions 
of  a  meter,  we  would  proceed  as  shown  in  find- 
ing the  area  of  a  square  (§  127).     Therefore, 
the  number  of  units  of  volume  in  a  rectangular  parallelepiped  is 
found  by  multiplying  the  number  of  units  of  area  in  the  base  by 
the  number  of  linear  units  in  the  height.     Or,  more  briefly,— 
Volume  of  a  rectangular  parallelepiped  =  base  X  height. 
If  a,  b,  and  c  denote  the  length,  the  breadth,  and  the  height 
of  the  solid,  then  the  base  =  a  X  bt  and  therefore  the  volume  = 
a  X  b  X  c. 

In  the  case  of  the  cube,  a  =  b  —  c ;  therefore,  — 
Volume  of  a  cube  =  a  X  a  X  a  =  a3. 

Here  a?  is  a  short  way  of  expressing  that  a  is  taken  as  a  factor 
three  times,  or  is  raised  to  the  third  power.  Hence  the  third  power 
of  a  number  is  often  called  the  cube  of  the  number. 

Remark.  —  Since  the  edge  of  a  cubic  meter  =  10  decimeters, 
therefore,  i  cubic  meter  =10X10X10—  1000  cubic  decime- 
ters (liters).  For  a  like  reason,  i  cubic  decimeter  =  1000  cubic 
centimeters.  In  general,  — 

The  ratio  between  two  units  of  volume  is  the  cube  of  the  ratio 
between  the  corresponding  units  of  length. 

Exercises.  —  Find  the  volumes  of  the  following  cubes :  — 

1.  Edge  =  8m.  3.    Edge  =  72cm.  5.   Area  of  base  =  36fim. 

2.  Edge  =  3.17™.          4.    Edge  =  9|in.  6.   Area  of  base  =  S. 


§   262.]     CHAPTER  XIII. SURFACES  AND  VOLUMES  OF  BODIES.  289 

Find  the  volumes  of  the  following  rectangular  parallelepipeds :  — 

7.  Length  =  4.5™,  breadth  =  7™,  height  =  3.4™. 

8.  Length  =  /,  breadth  =  b,  height  =  h. 

9.  Length  =  breadth  =  /,  height  =  h. 

10.  Area  of  base  =  8iim,  height  =  4™. 

11.  Area  of  base  =  S,  height  =  h. 

12.  How  many  cubes,  each  with  an  edge  — -  I*1™,  can  be  put  into  a  box 
I2dm  by  gdm  by  6dm,  inside  measurement  ? 

13.  Find  the  number  of  cubic  decimeters  of  wood  in  a  board  5™  long, 
o.6m  wide,  0.03™  thick.     What  kind  of  a  solid  is  the  board  ? 

14.  How  high  must  a  box  5dm  long  and  2dm  wide  be  to  hold  30  liters? 

15.  Volume  of  a  cube  =  74,o88ccm.     Find  the  edge. 

16.  Volume  of  a  cubical  tank  =  474.552ccm.     Find  its  depth. 

17.  Volume  of  a  cube  =  V.     Find  the  edge. 

IS.    Two  dimensions  of  a  rectangular   parallelepiped  are  9.37™  and  5m. 
Find  the  third  dimension. 

19.  A  room  is  8™  long,  6m  wide,  2.5"'  high.     How  many  cubic  meters  does 
it  contain  ?     How  much  more  space  would  it  inclose  if  the  floor  had  the  same 
perimeter,  but  were  in  the  shape  of  a  square  ? 

20.  How  many  stones  3dm  long,  i-5dm  wide,  O.75dm  thick,  are  required  to 
build  a  wall  40™  long,  3'"  high,  and  o.5m  thick  ? 

21.  How  many  liters  of  water  have  fallen  in  a  garden  42™  long  and  32m 
wide,  if  a  vessel  standing  in  it  is  filled  to  the  height  of  I2cm  ? 

22.  How  many  hogsheads  will  a  tank   lo*1  by  I2ft  by  9ft  hold  ?     (i  hogs- 
head =  63  gallons.) 

23.  Why  does  one  cubic  foot  contain  1728  cubic  inches. 

§  262.  THE  RIGHT  PARALLELOPIPED.  —  If  from  the  right  paral- 
lelepiped A  G  (Fig.  287)  we  take  away  the 


prism   B  G,  by  passing    through    the    edge  n^\            A/ 

BF  a  plane  BFMN  perpendicular  to  the  -fipf  -j -p^{ 

base,  and  then  place  this  prism  on  the  left 
in  the  position  A  H,  the  right  parallelepiped 

will  be  transformed  into  an  equivalent  rec-  |  /-;;-:'-" 

4-n  r**  m  1 1  or*  r^if  nll^l^-TiTT^ri  K  a  win  rr  tn  f*  en  VY\  P>   n  Pi  O"ni"  «^*                   " 


tangular  parallelepiped  having  the  same  height     A^  2? 

and  an  equivalent  base   (§   129).      There-  Fig.  287. 
fore,  by  §  261,  - 

Volume  of  a  right  parallelepiped  =  base  X  height. 


290 


GEOMETRY    FOR    BEGINNERS. 


[§   263. 


Since  area  ABCD  —  AB  X  B M,  therefore,  the  volume  of  the 
right  paraUelopipedAG  =  AB  X  B  MX  BF=  AB  X  BF  X  BM 
=  area  AB EFx  BM  =  a  rectangular  face  taken  as  base  x  the 
corresponding  height. 

Exercises.  —  1.  Find  the  volume  of  a  right  parallelepiped  if  the  height  = 
Iim,  and  the  base  is  a  parallelogram  having  the  length  I2m  and  the  altitude  5™. 

2.  A  wall  2Om  long,  3™  high,  is  o.81M  thick  at  the  bottom  and  0.4™  thick  at 
the  top.  Find  how  many  cubic  meters  it  contains. 

Hint.  —  Consider  the  wall  a  right  parallelepiped  having  for  base  a  trapezoid 
whose  parallel  sides  are  o.8m  and.o.4"1. 

§  263.  THE  OBLIQUE  PARALLELOPIPED.  —  Let  AG  (Fig.  288)  be 
an  oblique  parallelepiped.  Through  the  corner  E  pass  a  plane 
perpendicular  to  the  edge  EF,  cutting  from  the  parallelepiped  the 

solid  AMDNEHO,  and  place 
this  solid  on  the  right  in  the 
position  BPCQFGR.  By 
this  operation,  the  oblique  par- 
allelepiped is  transformed  into 
an  equivalent  right  parallele- 
piped, having  the  same  height, 
and  a  rectangular  base  EFRO 
equivalent  to  the  base  EFGH 
of  the  oblique  parallelepiped 
(§  129).  Now  the  volume  of 
the  right  parallelepiped  =  its  base  EFR  O  x  the  height  (§  262)  ; 
therefore,  — 

Volume  of  the  oblique  parallelepiped  =  base  EFGH  X  height. 

§  264.  THE  TRIANGULAR  PRISM.  —  If  through  the  edges  AB  and 
GH of  the  oblique  parallelepiped  AG  (Fig.  288)  we  pass  a  plane, 
it  will  divide  the  solid  into  two  oblique  triangular  prisms.  The 
same  plane  divides  the  equivalent  right  parallelepiped  into  two  right 
triangular  prisms.  These  right  prisms  are  equivalent,  each  to  its 


§   265.]     CHAPTER    XIII. — SURFACES  AND  VOLUMES  OF  BODIES.          291 

corresponding  oblique  prism,  because  each  is  formed  by  cutting 
off  (by  the  plane  through  E  _L  the  edge  EP)  a  portion  of  the 
oblique  prism,  and  transferring  it  to  the  right  of  the  figure.  Now 
these  right  prisms  are  also  equivalent  to  each  other ;  for,  if  one  of 
them  is  inverted,  it  is  then  clear  that  it  has  precisely  the  same 
shape  as  the  other,  and  that  its  faces  are  equal  and  parallel  to  those 
of  the  other.  Therefore,  the  two  oblique  prisms  are  equivalent, 
and  hence  each  is  half  of  the  parallelepiped.  Take  ADEN  as  the 
base  of  the  parallelepiped ;  then  the  distance  between  the  planes 
A D EH and  B  CFG  is  its  height,  the  two  prisms  have  the  same 
height,  and  their  bases  are  the  triangles  A  EH  and.  ADH,  each  of 
which  is  half  the  parallelogram  ADEN.  So  that  the  volume  of 
either  prism  (half  the  volume  of  the  parallelepiped)  is  equal  to  its 
base  (half  the  base  of  the  parallelepiped)  X  its  height.  In  general, — 

Volume  of  a  triangular  prism  =  its  base  X  its  height. 

Exercises.  —  Find  the  volume  of  a  regular  triangular  prism,  having  given,  — 

1.  Height  X  75cm,  area  of  base  =  i4Occm. 

2.  Height  —  h,  area  of  base  =  S. 

3.  Height  =  8.5m,  one  side  of  the  base  =  3™. 

4.  Height  =  h,  one  side  of  the  base  =  a. 

5.  The  bases  of  a  triangular  prism  8ocm  high  are  isosceles  right  triangles, 
a  leg  of  which  =  I2cm.     Find  the  volume  of  the  prism. 

6.  If  the  volume  of  a  triangular  prism  with  isosceles  right  triangles  for 
bases  =  6oocbm,  and  the  height  =  14™,  find  (z.)  area  of  the  base,  (zY.)  perime- 
ter of  the  base. 

§  265.  ANY  PRISM.  — Any  prism  (Pig.  289)  can  be  divided  into 
triangular  prisms  by  passing  planes,  as  shown  in  the 
figure,  through  one  of  the  lateral  edges.  The  trian- 
gular prisms  have  the  same  height  as  the  entire 
prisms,  and  the  sum  of  their  faces  is  equal  to  the 
base  of  the  entire  prism.  Therefore,  by  §  264  and 
by  addition,  it  follows  that  the  — 

Volume  of  any  prism  =  its  base  X  its  height. 

Hence,  any  two  prisms  having  equivalent  bases          Fig.  289. 
and  equal  heights  are  equivalent. 

•A 


292  GEOMETRY    FOR    BEGINNERS.  [§   266. 

Exercises.  —  Find  the  volumes  of  the  following  regular  prisms  :  — 

1.  Hexagonal  prism,  height  =  I7m,  side  of  base  =  2m. 

2.  Hexagonal  prism,  height  =:  h,  side  of  base  =  a. 

3.  Octagonal  prism,  height  =  85cm,  side  of  base  —  iocm 

4.  Octagonal  prism,  height  —  //,  side  of  base  =  a. 

§  266.  THE  CYLINDER.  —  If  we  inscribe  in  the  bases  of  a  cylin- 
der regular  polygons  of  the  same  number  of 
sides,  so  that  their  sides,  taken  pair  by  pair,  are 
parallel,  and  then  pass  planes  through  each  pair 
of  sides,  these  planes  with  the  polygons  will  form 
a  prism  inscribed  in  the  cylinder  (Fig.  290}. 
Now  the  greater  the  number  of  lateral  faces  in 
the  inscribed  prism,  the  nearer  does  its  volume 
approach  that  of  the  cylinder ;  that  is  to  say,  - 

Volume  of  cylinder  =  limit  of  volume  of  in- 
Fig.  290.     •       scribed  prism. 

The  volume  of  the  prism  =  its  base  X  its  height ; 
and  its  base  has  the  base  of  the  cylinder  for  its  limit.  Therefore, 
limit  of  the  volume  of  the  inscribed  prism  —  base  of  cylinder  x  its 
height. 

Hence  (Axiom  I.), — 

Volume  of  the  cylinder  =  its  base  X  its  height. 

Exercises.  — Find  the  volumes  of  the  following  cylinders  :  — 

1.  -Height  =  6m,  area  of  base  =  421™. 

2.  Height  =  8m,  radius  of  base  =  2m. 

3.  Height  =  64cm,  diameter  of  base  =  i6cm. 

4.  Height  =  7^,  circumference  of  base  =  44in. 

5.  Height  =  h,  radius  of  base  =  r. 

6.  Height  =  h,  radius  of  base  =  d. 

7.  The  volume  of  a  cylinder  =  2OOcbm,  the  height  =  13"'.     Find  the  radius 
of  the  base. 

8.  The  volume  of  a  cylinder  =  V,  the  height  =  h.     Find  the  radius  of  the 
base. 

{).    A  cylindrical  pail  holding  exactly  one  liter  is   i8cm  high.      Find  the 
diameter  of  the  base. 


§   267.]     CHAPTER  XIII. SURFACES  AND  VOLUMES  OF  BODIES. 


293 


10.  A  hollow  iron  tube  is  6m  long,  the  outer  diameter  =  92°™,  the  inner 
diameter  =  68cm.     How  many  cubic  decimeters  of  iron  are  there  ? 

11.  How  long  is  an  iron  tube,  if  the  inner  diameter  =  3.5cm,  the  outer  cir- 
cumference =  I5cm,  the  volume  =  i2Occm? 

12.  A  vessel  in  the  shape  of  a  cylinder  is  to  be  made  from  sheet-iron  2cm 
thick.     The  vessel  is  to  be  i'n  high,  and  to  hold  100  liters.     What  must  be  its 
outer  diameter  ? 


§  267.  THE  TRIANGULAR  PYRAMID.  —  Let  S-ABC  and  S 
-MNO  {Fig.  291)  be  two  triangular  pyramids  standing  on  th: 
same  horizontal  plane,  and  having  equivalent  bases  ABC  and 
MNO,  and  equal  heights.  Divide  the  common  height  into  any 
number  of  equal  parts,  and  through  the  points  of  division  pass 
planes  parallel  to  the  bases  of  the  pyramids.  Since  the  bases  are 
equivalent,  it  follows,  from  the  mode  of  generation  of  a  pyramid 
(§  246),  that  any  two  sections,  as  DEF  and  PQR,  situated  at 
the  same  distance  from  the  bases,  will  also  be  equivalent. 

In  the  two  pyramids  construct  triangular  prisms,  such  as  DEFGH1 
and  PQRTUV,  having  for  upper  base  one  of  the  sections;  for 
lateral  edges,  lines  parallel  to  SA  or  SM-,  and  for  a  common 
height,  the  distance  apart  of  two  sections.  It  follows,  from  §  265, 
that  these  prisms  will  be  equivalent,  taken  pair  by  pair,  in  the 
two  pyramids ;  therefore, 
the  sums  of  their  volumes 
will  be  equal,  no  matter 
how  many  or  how  few  of 
them  there  may  be.  But, 
the  greater  their  number 
the  nearer  the  sums  of 
their  volumes  approach  the 
volumes  of  the  two  pyra- 
mids ;  that  is  to  say,  — 

Volume  of  each  pyramid  =:  limit  of  sum  of  volumes  of  the  in- 
scribed prisms. 


N 


Fig.  291. 


294 


GEOMETRY    FOR    BEGINNERS. 


[§   267. 


Now  the  sums  of  these  volumes  remain  always  equal ;  therefore 
their  limits,  or  the  volumes  of  the  pyramids,  must  also  be  equal. 

Hence,   two  triangular  pyramids  having  equivalent  bases  and 
eqital  heights  are  equivalent. 

Now  let  S-ABC  (Fig.  292)  be  any  triangular  pyramid.     If 
we  pass  a  plane  through  BC  II  SA,  and  a  plane  through  S  II  ABC, 

we  form  a  triangular  prism 
A  B  CSP  Q.  This  prism  is 
divided  by  the  plane  SBC, 
and  a  plane  passed  through 
C  and  SP,  into  three  trian- 
gular pyramids,  S-ABC, 
P-  SB  C,  and  C  -  SPQ.  Of 

Fig.  292.  these,  S-AB  C  and  C-SPQ 

are  equivalent  because  they 

have  equivalent  bases  and  equal  heights.  Also,  P-SBC  and 
C-SPQ  are  equivalent,  because  we  may  take  *S  for  the  common 
vertex,  and  then  they  have  equal  heights  (the  distance  from  S 
to  the  plane  B  C  P  Q)  and  equivalent  bases  (the  equal  triangles 
PBC  and  PC  Q).  Therefore,  each  pyramid  =  £  the  prism.  Now 
the  volume  of  the  prism  =  base  x  height ;  therefore,  the  volume 
of  the  pyramid  S-ABC  (which  has  the  same  base  and  the  same 
height  as  the  prism)  =  %  base  X  height.  Therefore,  — 
Volume  of  a  triangular  pyramid '=  %  its  base  X  its  height. 

Exercises.  —  1.    Height  of  a  regular  triangular  pyramid—  I5m,  one  side 
of  the  base  =  2m.     Find  the  volume. 

2.  He.ight  of  a  regular  triangular  pyramid  =  h,  one  side  of  the  base  —-  a. 
Find  the  volume. 

3.  A  triangular  pyramid  9Ocm  high  has  an  isosceles  right  triangle  for  base, 
o«te  leg  of  which  —  4Ocm.     Find  the  volume  of  the  prism. 

4.  A  triangular  pyramid  with  an  isosceles  right  triangle  for  base  is  3.4™ 
high.     Its  volume  =  8ocm.     Find  (?.)  the  area  of  the  base  ;  (?V.)  the  perime- 
ter of  the  base. 

5.  The  frustum  of  a  regular  triangular  pyramid  is  4m  high,  and  two  homolo- 
gous sides  of  the  bases  are  2rn  and  i.25m.     Find  its  volume.     (See  §  249.) 


§    268.]     CHAPTER  XIII. SURFACES  AND  VOLUMES  OF  BODIES. 


295 


§  268.  ANY  PYRAMID.  —  Any  pyramid  whatever  can  be  divided 
into  triangular  pyramids  by  passing  planes  through 
one  of  the  lateral  edges,  as  shown  in  Fig.  293. 
These  triangular  pyramids  all  have  the  same  height 
as  the  entire  pyramid,  and  the  sum  of  their  bases 
=  the  base  of  the  entire  pyramid. 

Therefore,  —  . 

Volume  of  any  pyramid  =  -J  base  X  height.  Fig.  293. 

Exercises.  —  1.    Find  the  volume  of  a  regular  hexagonal  pyramid,  if  the 
height  =  36cm  and  a  side  of  the  base  =  8cm. 

2.  Find  the  volume  of  a  regular  octagonal  pyramid,  if  the  height  =  44"* 
and  a  side  of  the  base  =  4m. 

3.  Solve  Exercise  I,  if  the  height  =  h  and  a  side  of  the  base  =  a. 

4.  Solve  Exercise  2,  if  the  height  =  h  and  a  side  of  the  base  =  a. 

5.  The  frustum  of  a  regular  hexagonal  pyramid  is  2.7™  high,  and  two 
homologous  sides  of  the  bases  are  i.2m  and  7Ocm.     Find  the  volume. 

6.  A  vessel  has  the  shape  of  the  frustum  of  a  regular  four-sided  pyramid. 
It  is   i8cm  high,  and  the  outer  edges  of  its  bases  are  24cm  and  i6cm.     How 
many  liters  of  water  will  it  hold  if  the  material  of  which  it  is  made  is  2cm  thick  ? 


§  269.  THE  CONE.  —  Inscribe  in  the  base  of  a  cone  (Fig. 
a  regular  polygon,  and  pass  planes  through  the 
sides  of  this  polygon  and  the  vertex  of  the  cone. 
These  planes  with  the  polygon  form  an  in- 
scribed regular  pyramid  whose  volume  ap- 
proaches the  nearer  to  that  of  the  cone,  the 
greater  the  number  of  sides  in  the  regular  poly- 
gon ;  that  is,  — 

Volume  of  cone  =  limit  of  volume  of  inscribed 
pyramid. 

But  volume  of  any  pyramid  =  £  its  base  X  its 
height.     And  in  this  case  the  base  has  the  base 
of  the  cone  for  its  limit.     Therefore,  limit  of  volume  of  inscribed 
pyramid  =  •£  base  of  cone  X  height.     Therefore  (Axiom  I.), — 

Volume  of  a  cone  =  %  its  base  X  its  height. 


Fig.  294. 


296  GEOMETRY    FOR    BEGINNERS.  [§   270. 

Exercises.  —  Find  the  volumes  of  the  following  cones  :  — 

1.  Height  =  15™,  area  of  base  =  6ocim. 

2.  Height  =  I2rn,  radius  of  base  ~  3m. 

3.  Height  =  h,  radius  of  base  =  r. 

4.  Height  —  74cm,  diameter  of  base  =  4.6™. 

5.  Height  —  h,  diameter  of  base  =  d. 

6.  Height  —  6ft,  circumference  of  base  =  ijjft. 

7.  Slant  height  =  5™,  radius  of  base  =  3™. 

8.  Slant  height  =  k,  radius  of  base  —  r. 

9.  Find  the  volume  of  a  cone  whose  slant  height  =  diameter  of  the  base 

-  2™. 

10.  Volume  of  a  cone  —  8oocbm,  radius  of  base  —  i.8m.     Find  the  height. 

11.  Find  the  volume  of  the  frustum  of  a  cone,  if  the  radii  of  the  bases  are 
7-5cm  and  5cm,  and  the  height  25cm.     (See  §  249.) 

12.  A  vessel  has  the  shape  of  the  frustum  of  a  cone  im  high,  and  the  inner 
circumferences  of  the  bases  are  5™  and  4m.     How  many  cubic  meters  of  water 
will  it  hold  ? 

13.  What  part  of  a  cone  remains  if  a  cone  two-fifths  as  high  as  the  entire 
cone  is  cut  off  ? 

14.  How  many  gallons  of  water  will  a  vessel  hold  which  has  the  shape 
of  the  frustum  of  a  cone,  the  height  being  3ft,  the  diameters  of  the  bases  i8in 
and  I2in  ? 

15.  A  Dutch  windmill  in  the  shape  of  the  frustum  of  a  cone  is  I2m  high. 
The  outer  diameter  at  the  bottom  and  the  top  are  i6m  and  I2m  ;   the  inner 
diameters  I2m  and  iom.     How  many  cubic  meters  of  stone  were  required  to 
build  it  ? 

16.  The  chimney  of  a  factory  is  32™  high,  and  the  outer  circumferences  of 
its  bottom  and  top  are  20™  and  I4m,  while  the  bore  is,  throughout,  2m  wide. 
How  many  cubic  meters  of  brick  does  it  contain  ? 

§  270.   The  results  obtained  in  H  261-266  may  be  summed 
up  in  one  general  rule,  as  follows  :  — • 

Volume  of  a  prism  or  a  cylinder  =  base  X  height. 

And  the  results  obtained  in  §  $  267-269  may  be  summed  up  in 
one  general  rule,  as  follows  :  — 

Volume  of  a  pyramid  or  a  cone  =  J  base  x  height. 


§   271.]     CHAPTER  XIII. SURFACES  AND  VOLUMES  OF  BODIES.  297 

§  271.  THE  SPHERE.  —  In  order  to  find  the  volume  of  a  sphere, 
conceive  a  series  of  pyramids  formed,  having  for  a  common  vertex 
the  centre  of  the  sphere,  and  for  bases  the  faces  of  a  polyhedron 
circumscribed  about  the  sphere  (that  is,  a  polyhedron  whose  faces 
touch  the  surface  of  the  sphere,  each  in  a  single  point).  By  in- 
creasing the  number  of  these  pyramids,  we  can  make  their  sum 
approach  the  volume  of  the  sphere  as  nearly  as  we  please ;  or,  in 
other  words,  — 

Volume  of  sphere  =  limit  of  sum  of  the  pyramids. 

Now  each  pyramid  =  £  height  X  base,  and  their  common  height 
=  radius  of  the  sphere  ;  therefore,  sum  of  the  pyramids  =  £  radius 
of  sphere  x  sum  of  the  faces.  The  limit  of  the  sum  of  the  faces 
(when  their  number  is  increased  more  and  more)  =  the  surface  of 
the  sphere.  Therefore,  limit  of  the  sum  of  the  pyramids  =  £  radius 
X  surface  of  the  sphere.  Therefore  (Axiom  I.), — 

Volume  of  a  sphere  ==  £  radius  X  surface  of  sphere. 

Exercises.  — 1.    Find  the  volume  of  a  sphere  if  the  radius  =  6m. 

2.  Find  the  volume  of  a  sphere  if  the  radius  =  r. 

3.  Find  the  volume  of  a  sphere  if  the  diameter  —  14°™. 

4.  Find  the  volume  of  a  sphere  if  the  diameter  =  d. 

5.  Find  the  volume  of  a  sphere  if  the  circumference  of  a  great  circle  =  ioft. 

6.  Required  the  volume  of  the  largest  sphere  that  can  be  turned  from  a 
cubical  block  of  wood  whose  edge  =  idm.     How  much  of  the  wood  is  lost  ? 

7.  The  volume  of  a  sphere  =  i8ocbm.     Find  the  radius. 

8.  The  volume  of  a  sphere  =  V.     Find  the  radius. 

9.  Find  the  volume  of  the  largest  sphere  that  can  be  turned  from  a  cylin- 
der of  wood  whose  diameter  =  its  height  —  I2cm. 

10.  The  diameter  of  a  cylinder  =  its  height  =  h.     Compare  the  volume  of 
this  cylinder  with  that  of  the  largest  sphere  that  can  be  inclosed  in  it. 

Ans.   Volume  of  the  sphere  =  f  volume  of  the  cylinder. 
NOTE.  —  This  curious  fact  was  discovered  by  Archimedes  about  230  B.C. 

11.  Find  the  difference  between  the  volumes  of  two  concentric  spheres 
whose  radii  are  4m  and  5™. 

12.  In  two  concentric  spheres,  the  volume  of  the  larger  =  twice  that  of  the 
smaller.     If  the  radius  of  the  smaller  =  im,  find  the  radius  of  the  larger. 


298  GEOMETRY   FOR   BEGINNERS.  [§   272. 

§  272.  The  volume  of  any  body,  whatever  be  its  shape,  may  be 
found  by  weighing  it,  and  then  dividing  the  weight  by  the  weight 
of  one  unit  of  volume.  The  quotient  will  be  the  number  of  units 
of  volume  in  the  body. 

In  the  metric  system,  the  most  common  units  of  weight  are, — 

The  GRAM  (g.)=  the  weight  of iccm  of  pure  water  (at  4°  C.). 
The  KILOGRAM  (kg.)=  weight  of  icdm  of  pure  water  (at  4°  C.). 

Since  icdm  =  ioooccm,  i  kilogram  =  1000  grams. 

The  ratio  of  the  weight  of  any  substance  to  the  weight  of  the 
same  volume  of  pure  water  is  called  the  SPECIFIC  GRAVITY  of  that 
substance. 

SPECIFIC  GRAVITIES  OF  SOME  COMMON  SUBSTANCES. 


Platinum     . 

.      22.00 

Brass  (sheet)     . 

8.40 

Boxwood 

.       1.20 

Gold  . 

IQ  35 

Steel    .... 

780 

Ice       ... 

O  Q2 

Mercury 

.       13.60 

Iron  (cast)    .     . 

/  •*** 

7-25 

Oak     .     .     . 

.      0.90 

Lead       .     . 

•     "-35 

Zinc     .... 

7.19 

Alcohol    .     . 

.      0.80 

Silver      .     . 

.     10.51 

Marble     .     .     . 

2.70 

Pine     .     .     . 

•    o-57 

Copper  .     . 

.      8.80 

Glass    .... 

2.60 

Cork    .     .    '. 

.    0.24 

The  gram  and  the  kilogram  being  by  definition  the  weights  of 
unit  volumes  of  water,  it  follows  that  the  above  numbers  express 
the  weights  of  iccra  of  the  substance  in  grams,  or  of  icdm  (i  liter)  of 
the  substance  in  kilograms. 

Whence  it  is  clear  that  in  the  metric  system,  — 


T7  /          *     t  j 

Volume  of  a  body  =  _  — 

its  specific  gravity 

the  volume  being  expressed  (/.)  in  cubic  centimeters,  or  (it.)  in 
cubic  decimeters  (liters),  according  as  the  weight  is  expressed 
(*'.)  in  grams  or  («'.)  in  kilograms. 

The  most  important  English  units  of  weight  are  the  POUND  (Ib.) 
avoirdupois,  the  OUNCE  (oz.)  =  jV1*,  and  the  TON  =  2Ooolb. 


§   272.]     CHAPTER  XIII.  —  SURFACES  AND  VOLUMES  OF  BODIES.          299 

They  are  connected  (nearly  enough  for  practical  purposes)  with 
the  weight  of  unit  volume  of  water  by  the  relation,  — 

Weight  of  i  cubic  foot  of 'water =  62^lbs  =  iooooz. 

The  weight  in  pounds  of  a  cubic  foot  of  any  other  substance  is 
found  by  multiplying  its  specific  gravity  by  62^,  and  the  weight  in 
ounces  by  multiplying  its  specific  gravity  by  1000. 

NOTE.  —  283g  — T.OZ  (nearly);  iks  =  2.2^3  (nearly). 

Exercises.  —  1.    Find  the  volume  of  a  piece  of  granite  which  weighs  4ks. 

2.  A  piece  of  lead  weighs  24ks.     Find  its  volume. 

3.  What  is  the  volume  of  848  of  silver  ? 

4.  Required  the  edge  of  a  cubical  vessel  which  will  hold  2OOOlbs  (i  ton) 
of  mercury. 

5.  Find  the  edge  of  a  marble  cube  weighing  184^. 

6.  An  iron  cylinder  2.$m  long  weighs  68oks.     Find  its  diameter. 

7.  Find  the  edge  of  a  cube  of  cork  which  weighs  the  same  as  an  iron  ball 
22cm  in  diameter. 

8.  How  many  liters  will  a  vessel  hold  if  the  vessel  weighs  when  empty 
I.5kS,  and  when  full  of  water  i-5ks  ? 

9»    An  oak  cone  40°™  high  weighs  2ks.     Find  its  diameter. 

10.  What  is  the  diameter  of  a  cannon  ball  which  weighs  64lbs  ? 

11.  How  can  the  weight  of  a  body  be  found  if  its  volume  and  its  specific 
gravity  are  known  ? 

12.  Find  the  weight  of  IM  of  water,  ihl  of  alcohol,  and  icubft  of  mercury. 

13.  Find  the  weight  of  a  rectangular  parallelepiped  of  copper,  o.34m  long, 
o.nm  wide,  o.O3m  thick. 

14.  Find  the  weight  of  an  oak  cone,  if  the  radius  of  the  base  —  32cm  and 
the  height  =  9Ocm. 

15.  Find  the  weight  of  a  hollow  brass  cube,  if  the    inner  diameter  =  ^ 
and  the  thickness  of  the  brass  =  icm. 

16.  If  a  cube,  whose  edge  =  8cm,    weighs   508,    find    the  weight  of  the 
largest  cylinder  that  can  be  turned  from  it.      Also  the  weight  of  the  largest 
cone.     Also  the  weight  of  the  largest  sphere. 

17.  How  can  the  specific  gravity  of  a  body  be  found,  if  its  weight  and  its 
volume  are  known  ? 

18.  A  cubical  block  of  pine  wood,  an  edge  of  which  —  I2cm,  weighs  ik£. 
Find  the  specific  gravity  of  the  wood. 

19.  If  a  mass  of  ice  containing  27Ocbm  is  known  to  weigh  229,000^,  find 
the  specific  gravity  of  ice. 

20.  Find  the  specific  gravity  of  cast  iron,  if  2^cutt  weigh  953lbs. 


300  GEOMETRY    FOR    BEGINNERS,  [§   273. 

§  273.  Equivalent  solids  have  the  same  volume  (size)  ;  similar 
solids,  the  same  shape ;  equal  solids,  the  same  volume  and  the 
same  shape. 

•     Equal  polyhedrons  must  fulfil  three  conditions  :  — 

(/.)  They  must  have  the  same  number  of  faces,  equal  each  to 
each. 

(i/.)    Their  homologous  edges  must  be  equal. 

(m.)    Their  homologous  solid  angles  must  be  equal. 

Similar  polyhedrons  must  also  fulfil  three  conditions  :  — 

(/.)  They  must  have  the  same  number  of  faces  similar  each  to 
each. 

(«'.)    Their  homologous  edges  must  be  proportional. 

(iii.)    Their  homologous  solid  angles  must  be  equal. 

In  both  cases,  there  are  the  same  number  of  edges  and  also  of 
corners. 

Conversely,  if  all  three  of  the  above  conditions  hold  true  of  two 
polyhedrons,  then  these  polyhedrons  are  equal  or  similar,  as  the 
case  may  be. 

There  are  four  important  classes  of  similar  solids  :  — 
(/.)    All  cubes  and  regular  polyhedrons  of  the  same  name  are 
similar. 

(//.)  In  order  that  two  regular  prisms  or  two  regular  pyramids 
may  be  similar,  they  must  have  the  same  number  of  faces,  and  their 
homologous  lateral  edges  must  be  proportional  to  the  homologous 
sides  of  their  bases.  But  the  homologous  lateral  edges  in  the 
prisms  are  obviously  proportional  to  the  heights,  and  in  the  pyra- 
mids they  are  easily  proved  to  be  proportional  to  the  heights,  by 
reasoning  as  in  §  249 ;  and  the  homologous  sides  of  the  bases  are 
(by  §  167)  proportional  to  the  less  radii  of  the  bases.  Therefore, 
two  regular  prisms  or  pyramids  with  the  same  number  of  sides  are 
similar,  if  their  heights  are  proportional  to  the  less  radii  of  their 
bases. 


§  274-]    CHAPTER  xm.  — SURFACES  AND  VOLUMES  OF  BODIES.       301 

(in.)  By  regarding  the  cylinder  or  the  cone  as  the  limit  of  the 
inscribed  regular  prism  or  regular  pyramid,  when  the  number  of 
its  sides  is  increased  more  and  more  (§§  266,  269),  it  becomes 
evident  that  two  cylinders  or  two  cones  are  similar,  if  their  heights 
ire  proportional  to  the  radii  of  their  bases. 

(iv.)    All  spheres  are  similar. 

§  274.  If  the  edge  of  a  cube  =  a,  the  (total)  surface  =  6  a2,  and 
the  volume  —  a3. 

Now,  if  we  have  other  cubes  whose  edges  are  2  a,  3  a,  4  a,  etc., 
their  surfaces  will  be  4,  9,  16,  etc.,  times  that  of  the  first  cube,  since 
(2tf)2  =  4#2,  (3<?)2  =  9#2,  (4#)2  =  i6a2,  etc.;  and  their  volumes 
will  be  8,  27,  64,  etc.,  times  that  of  the  first  cube,  since  (2  a)s  =  803, 
(3^)3=  27#3,  (4«)s  =  64#3,  etc. 

Bearing  in  mind  that  4,  9,  16,  etc.,  are  the  sqttares,  and  8,  27, 
64,  etc.,  the  cubes,  of  2,  3,  4,  etc.,  we  see  that  in  cubes  the  surface 
is  proportional  to  the  square,  and  the  volume  proportional  to  the 
cube,  of  the  edge. 

What  has  here  been  shown  to  be  true  of  the  surface  and  volume 
of  the  cube  in  respect  to  its  edges,  can  be  shown  to  be  true  of 
the  surface  and  volume  of  all  similar  solids  in  respect  to  any 
homologous  lines  (edges,  heights,  radii,  diameters,  etc.). 

That  is  to  say,  — 

1.  The  surfaces  of  two  similar  solids  are  proportional  to  the 
SQUARES  of  any  two  homologotis  lines. 

II.  The  volumes  of  two  similar  solids  are  proportional  to  the 
CUBES  of  any  two  homologous  lines. 

Exercises.  —  1.  Compare  the  surface  and  the  volume  of  a  cube  whose 
edge  =  2m  with  those  of  a  cube  whose  edge  =  im. 

2.  The  edge  of  a  cube  =  im.     Find  the  edge  of  a  cube,  (V.)  having  twice 
the  surface,  («')  having  twice  the  volume  of  the  first  cube. 

3.  Required  the  depth  of  a  cubical  tank  which  will  hold  64  times  as  much 
as  a  similar  tank  whose  depth  =  3m. 


302  GEOMETRY    FOR    BEGINNERS.  [§   274. 

.     4.    Compare    (i.)    the    surfaces,    (zY.)    the    volumes,    of   the    rectangular 
parallelepipeds  whose  edges  are  a,  b,  c,  and  3 a,  3 b,  3  £. 

5.  Given  two  similar  cylinders  whose  diameters  are  d  and  4^,  and  whose 
heights  are  h  and  4/2.    Prove  that  Laws  I.  and  II.,  stated  above  in  italics,  hold 
true  of  them. 

6.  Same  exercise  applied  to  two  similar  cones. 

7 .  Given  two  spheres  with  radii  r  and  5  r.     Prove  that  the  above  laws  hold 
true  of  their  surfaces  and  their  volumes. 

8.  The  edge  of  a  polyhedron  =  8m,  and  the  homologous  edge  of  a  similar 
polyhedron  =  5m.     How  many  times  will  the   first   polyhedron   contain  the 
second  ? 

9.  A  box  is  9Ocm  long,  58cm  wide,  and  43cm  deep.    Find  the  dimensions  of 
a  similar  box  twice  as  large. 

10.  Find  the  volume  of  the  box  in  the  last  exercise  if  its  three  dimensions 
are  doubled. 

11.  Find^its  volume  if  one  dimension  is  doubled.     If  two  dimensions  are 
doubled. 

12.  I  wish  to  make  an  iron  cylinder,  similar  to  a  cylinder  whose  diameter 
=  26cm  and  whose  height  =  64cm,  but  three  times  as  large.     What  must  be 
its  dimensions  ?     How  many  kilograms  of  iron  are  required  ? 

13.  How  far  from  the  vertex  of  a  cone  5om  high  must  a  plane  parallel  to 
the  base  be  passed  in  order  that  the  cone  cut  off  may  be  exactly  y^Vo  °f  tne 
entire  cone  ? 

14.  Divide  into  two  equivalent  parts,  by  a  plane  parallel  to  the  base,  a 
cone  65cm  high,  and  having  a  base  whose  diameter  =  30°™. 

15.  If  the  radius  of  a  sphere  =  37cm,  what  will  be  the  radius  of  a  sphere 
ten  times  as  large  ? 

16.  I  have  a  cylindrical  boiler  6oft  long  and  8ft  in  diameter,  and  I  wish 
another  boiler  similar  in  shape  but  f  as  large.    What  must  be  its  dimensions  ? 

17.  If  the  edge  of  a  regular  tetahedron  =  im,  it  can  be  proved  that  the 
surface  =  1.732^™,  and  the  volume  =  o.i  I78cbm;  find  the  surface  and  the  volume 
if  the  edge  =  7™. 

18.  If  the  edge  of  a  regular  octahedron  -  im,  it  can  be  proved  that  the 
surface  =  3.4641  lm,  and   the  volume  =  0.471 4cbm;    find  the   surface  and  the 
volume  if  the  edge  =  4cm. 


§275-]     CHAPTER  XIII.  —  SURFACES  AND  VOLUMES  OF  BODIES.  303 

III.  —  Exercises  and  Applications. 
§  275.   PRISMS  AND  CYLINDERS. 

1.  How  many  steres  of  wood  in  a  cubical  pile  one  edge  of  which  =  I2m  ? 

2.  Total  surface  of  a  cube  =  8o*icm;  find  its  volume. 

3.  Total  surface  of  a  cube  =  5;  find  its  volume. 

4.  Volume  of  a  cube  —  2OOcbm;  find  its  surface. 

5.  Volume  of  a  cube  =  V;  find  its  surface. 

6.  A  diagonal  on  one  face  of  a  cube  =  I2cm;  find  the  volume  of  the  cube. 

7.  Edge  of  a  cube  =  4cm;  find  the  diagonal  through  the  cube. 

8.  Edge  of  a  cube  =  a  ;  find  the  diagonal  through  the  cube. 

9.  A  diagonal  through  a  cube  =  3Ocm;  find  the  edge. 

10.  Find  the  edge  of  a  cork  cube  which  will  weigh  the  same  as  lccm  of 
iron.     (See  Table,  page  298.) 

11.  Volumes  of  two  cubes  are  as  8  :  27;  what  is  the  ratio  of  their  edges  ? 

12.  Find  the  edge  of  a  cube  half  as  large  as  a  cube  whose  edge  =  i6cm. 

13.  Edges  of  two  cubes  are  6cm  and  I2cm;   find  the  edge  of  a  cube  equal 
to  their  sum. 

14-.    Edges  of  two  cubes  are  I2cm  and  i6cm;  find  the  edge  of  a  cube  equal 
to  their  difference. 

15.  Find  the  edge  of  a  cubical  tank  which  holds  looo111  of  water. 

16.  A  water  tank  has  a  rectangular  base  3.2m  by  2.5™.     How  many  liters 
of  water  does  it  contain  when  the  height  of  the  water—  i.24m? 

17.  A  tank  has  a  square  base  one  edge  of  which  =  3m.     How  deep  is  the 
water  when  the  tank  contains  iSoo1? 

18.  A  cubical  reservoir  whose  edge  —  24™  contains  how  many  hectoliters 
of  water  when  the  water  is  4™  deep  ? 

19.-  If  water  enters  the  above  reservoir,  through  a  tube,  at  the  rate  of  I6O1 
per  minute,  in  what  time  will  it  be  full  ? 

20.  If  it  is  found  that  in  24  hours  the  reservoir  loses  by  evaporation  a 
layer  2cm  in  depth,  how  many  liters  does  it  lose  ? 

21.  What  time  will  be  required  to  fill  the  reservoir,  taking  account  of 
evaporation  ? 

22.  How  large  a  field  can  be  irrigated  by  the  reservoir  full  of  water  if 
8oocbm  of  water  are  required  for  every  hectare  ? 

23.  If  the  dimensions  of  a  rectangular  parallelepiped  are  a,  If,  and  c,  find 
the  length  of  the  diagonal  from  an  upper  corner  to  the  lower  opposite  corner  ? 


304  GEOMETRY    FOR    BEGINNERS.  [§   275. 

24.  What  is  the  cost  of  digging  a  cellar  n.5m  long,  8.8m  wide,  and  2.62m 
deep,  at  the  rate  of  $0.80  per  cubic  meter  ? 

25.  What  is  the  weight  of  a  marble  block  2.75m  long,  o.35m  wide,  and 
2.25™  thick  ? 

26.  A  body  weighs  less  in  water  than  in  air  by  exactly  the  weight  of  its 
pwn  volume  of  water.     If  a  body  i.2m  long,  o.62m  wide,  and  0.4™  thick  weighs 
4Ook»  in  air,  what  will  it  weigh  in  water  ? 

27.  Find  the  specific  gravity  of  the  body  in  the  last  exercise. 

28.  How  much  will  a  block  of  marble  2m  long,  i.6m  wide,  and  o.8m  thick 
weigh  under  water  ? 

29.  Which  will  weigh  the  most  in  water,  an  iron  cube  whose  edge  =  I2cm 
or  a  rectangular  parallelepiped  of  lead  I5cm  long,  iocm  wide,  and  8cm  thick  ? 

30.  In  a  room  6.2m  long,  4.75™  wide,  and  2.88m  high  are  6  persons.     In 
what  time  will  the  air  in  the  room  become  unfit  for  respiration  if  each  person 
breathes  20  times  a  minute,  and  at  each  breath  vitiates  4OOccm  of  air  ? 

31  •    What  is  the  weight  of  the  air  in  the  above-mentioned  room  ?   (Specific 
gravity  of  the  air  =  0.0013.) 

32.  What  must  be  the  height  of  a  box  2.5m  long  and  2m  wide,  that  it  may 
hold  22.5cbm  ? 

33.  A  rectangular  vessel  48cm  long  and  36cm  wide  contains  56  liters  of 
water.     What  is  the  depth  of  the  water  ? 

34.  Into  a  corn-bin  8m  long  and  5.2™  wide,  56hl  of  corn  are  put;  find  the 
height. 

35.  The  total  surface  of  a  cube  —  54km;  find  the  sum  of  its  edges. 

36.  Ten  leaden  cubes,  each  with  an  edge  —  2Ocm,  are  melted  into  one 
cube;  find  its  edge. 

37.  If  the  dimensions  of  a  trunk  are  im,  I5m,  and  2m,  find  the  dimensions 
of  a  similar  trunk  that  will  hold  eight  times  as  much. 

38.  Find  the  dimensions  of  a  similar  trunk  holding  twice  as  much. 

39.  A  wooden  trough  is  to  be  made,  whose  inner  dimensions  are  to  be, 
length  3.35™,  breadth  i.6m,  depth  im  ;   the  ends  are  to  be  iocm  thick,  the  sides 
8cm,  and  the  bottom  I2cm.    (?'.)  How  many  cubic  metres  of  wood  are  required  ? 
(zY.)   How  many  liters  will  the  trough  hold?     If  the  trough  is  made  of  oak 
wood,  what  will  it  weigh?  (iii.')  when  empty?  (iv?)  when  full  of  water? 

40.  What  will  it  cost  to  dig  a  ditch  around  a  field  in  the  shape  of  a  square, 
containing  4  hectares,  if  the  ditch  is  to  be  2m  deep,  im  wide  at  the  bottom, 
i.4m  at  the  top,  and  the  price  is  $0.25  per  cubic  metre  of  earth  thrown  out? 

41.  How  many  cubic  metres  of  manure  are  required  to  cover  with  a  layer 
2cm  thick  a  triangular  field  whose  base  =  840'"  and  altitude  =  320"*? 


§275-]     CHAPTER  XIII.  -  SURFACES  AND  VOLUMES  OF  BODIES.  305 

42.  If  78oohl  of  gravel  are  spread  upon  a  rectangular  meadow  560™  by 
280™,  how  much  gravel  is  required  to  make  a  layer  of  the  same  thickness 
upon  a  trapezoidal  field  whose  parallel  sides  are  920™  and  74Om,  and  alti- 
tude 250™? 

43.  Through  an  orifice  in  the  side  of  a  vessel  6cm  square,  water  flows  at 
the  rate  of  i6m  per  second.      How  many  hectoliters  will  flow  out  in  one 
minute  ? 

44.  Through  a  rectangular  orifice  I2cm  by  8cm  water  flows  at  the  rate  of 
2om  per  second.     How  much  water  will  flow  out  in  one  hour? 

45.  If  one  bushel  of  wheat  makes  48  Ibs.  of  flour,  and  one  barrel  of  flour 
contains  196  Ibs.,  how  many  barrels  of  flour  can  be  made  from  the  contents  of 
a  bin  2Oft  long,  ioft  wide,  and  6ft  deep,  filled  with  wheat  ? 

46.  How  many  tons  of  ice  can  be  packed  in  a  building  60**  long,  40**  wide, 
and  30*"'  high  ? 

47.  How  many  tons  of  coal  will  a  bin  2Oft  long,  i6ft  wide,  and  I2ft  deep 
contain,  allowing  4Ocub  ft  to  a  ton  ? 

48.  How  many  gallons  of  water  will  a  cistern  8ft  deep,  and  6ft  square  at 
the  bottom,  hold  ? 

49.  A  merchant  imports  4Ohl  of  wine.     What  must  be  the  dimensions,  in 
feet,  of  a  cubical  tank  which  will  just  hold  it  ? 

50.  After  a  shower,  I  find  a  dish,  placed  in  the  open  air,  filled  with  water 
to  the  depth  of  half  an  inch.     How  many  tons  of  water  have  fallen  upon  one 
square  mile,  the  fall  being  supposed  everywhere  the  same  ? 

51.  How  many  bricks  8in  X  4in  X  2in  will  be  required  to  build  a  wall  50** 
long,  3oft  high,  and  i6|in  thick,  allowing  £in  in  each  dimension  for  the  mortar? 

52.  How  many  cords  of  wood  in  a  pile  60**  long,  8ft  wide,  and  7ft  high  ? 
NOTE.  —  One  cord  of  wood  is  a  pile  8ft  long,  4ft  wide,  and  4ft  high. 

53.  A  stick  of  timber  is  I4in  X  18^  at  the  end.     What  length  of  it  will 
contain  30  cubic  feet  ? 

54.  Find  the  contents,  in  board  feet,  of  a  board  i8ft  long,  I5in  wide,  and 
2in  thick  ? 

NOTE.  —  One  board  foot  is  ift  long,  ift  wide,  and  iin  thick. 

55.  Find  the  cost  of  20  planks,  each  i6ft  long,  i8in  wide,  and  3in  thick,  at 
$2.00  per  hundred  feet,  board  measure. 

56.  How  many  feet  of  lumber,  board  measure,  will  be  required  to  make  a 
box  6ft  X  4ft  X  3ft,  outside  measurement,  the  thickness  of  the  lumber  to  be 


57.    Find  the  volume  of  a  3-sided  regular  prism  if  a  lateral  edge  =  a  side 
of  the  base  =  24°™. 


306  GEOMETRY    FOR    BEGINNERS.  [§   275. 

58.  In  a  regular  hexagonal  prism  the  height  =  3  times  the  less  radius  of 
the  base  =  6ocm.     Find  the  height  of  a  similar  prism,  made  of  glass,  which 
weighs  6ks. 

59.  How  high  must  a  regular  hexagonal  prism  be  to  hold  one  hectoliter  if 
one  side  of  the  base  =  64°™  ? 

60.  A  cylindrical  vessel,  the  area  of  whose  base  =  51™,  contains  75hl  of 
water  when  full ;   find  the  height  of  the  vessel. 

61.  Give  the  dimensions  of  a  cylindrical  vessel  which  will  hold  1000  liters. 
Is  this  question  determinate  or  indeterminate?     Why? 

62.  Required  the  dimensions  of  a  similar  cylinder  that  will  hold  ^  as  much? 

63.  If  a  vessel  holding  one  hectoliter  is  a  cylinder  whose  height  is  equal 
to  the  diameter  of  the  base,  find  the  height. 

64.  A  cylindrical  cistern  3451"  deep  and   i.im  wide  is  to  be  lined  with 
stones,  each   of  which,  mortar  included,  measures   I2cm  by  6cm.     How  many 
stones  are  required? 

65.  A  cylindrical  boiler  is  58m  long  and  has  a  diameter  of  0.9™.     Find  the 
area  of  the  convex  surface  exposed  to  the  fire,  if  this  surface  intersects  the 
circular  end  of  the  boiler  in  an  arc  of  130°. 

66.  If  a  glass  cylinder  o.35m  in  diameter  and  im  in  height  is  filled  with 
water,  and  the  water  is  then  poured  into  another  cylinder  o.8in  in  diameter, 
how  high  will  the  water  rise  in  the  latter  vessel? 

67 .  Out  of  a  cylindrical  reservoir  8.8m  in  diameter  water  flows  at  the  rate 
of  21  per  second.     Through  what  distance  will  the  surface  of  the  water  fall  in 
one  hour? 

68.  About  the  convex  surface  of  a  cylinder  o.8m  in  diameter  and  i.2m  high, 
a  cord  2mm  in  diameter  is  wound  until  the  surface  is  completely  covered. 
How  many  meters  of  cord  are  required? 

69.  A  well  is  to  be  I5m  deep  and  2. 3™  in  diameter.     The  price  for  digging 
it  is  to  be  $0.40  per  cubic  meter  for  the  first  three  meters  of  depth,  $0.25  more 
for  the  next  three  meters,  and  so  on,  $0.25  being  added  to  the  price  for  each 
successive  three  meters  of  depth.     Find  the  cost  of  digging  the  well. 

70.  A  cast-iron  cylinder  3™  long  and  2Ocm  in  diameter  is   reduced  in  a 
turning  lathe  to  a  diameter  of  i8cm.     Find  the  loss  in  weight. 

71.  The  horizontal  section  of  a  bath-tub  im  high  is  an  ellipse  whose  axes 
are  i.8m  and  o.64m.     How  many  liters  of  water  will  it  hold?     (See  §  221.) 

72.  How  large  a  cylinder  can  be  made  by  rolling  up  a  rectangular  sheet 
of  zinc  8ocm  by  6ocm,  if  the  height  of  the  cylinder  is  8o(;m  ?     How  large  will 
the  cylinder  be  if  the  height  is  6ocm  ? 

73.  An  iron  cylinder  8ocm  long  and  weighing  56ks  is  bored  through  the 
centre  until  £  of  the  iron  is  removed;  find  the  thickness  of  what  is  left. 


§   276.]     CHAPTER   XIII. — SURFACES  AND  VOLUMES  OF  BODIES.         307 


74.  A  cylindrical  vessel  4™  high  and  i.6m  in  diameter  is  half-full  of  water. 
Through  an  orifice  at  the  bottom  4cm  in  diameter  the  water  is  running  out  at 
the  rate  of  6m  a  second,  and  through  a  pipe  2cm  in  diameter  it  is  running  into 
the  vessel  at  the  rate  of  I2m  a  second;   find  the  level  of  the  water  after  one- 
quarter  of  an  hour. 

75.  What  must  be  the  diameter  of  a  supply-pipe,  for  the  vessel  in  the  last 
exercise,  which  will  supply  water  as  fast  as  it  runs  out?  0 

76.  What  must  be  its  diameter  in  order  that  the  vessel  may  be  filled  in 
ten  minutes? 

77.  What  must  be  its  diameter  in  order 
that  the  vessel  may  be  emptied  in  one  hour? 

78.  The  piston  of  a  pump  is  36°™  in  diam- 
eter,  and   moves  through    a   space  of    5Ocm. 
How  many  liters  of  water  are  thrown  out  by 
100  strokes? 

79.  When  a  body  is  placed  under  water, 
in  a  cylinder  6ocm  in  diameter,  the  level  of  the 
water  is  observed  to  rise  30°°  {Fig.  295)  ;  find 
the  volume  of  the  body. 

80.  How  much  will  a  brass  cylinder  weigh 
under  water  if  the  height  =  64cm  and  the  diam- 
eter of  the  base  =  40°™?     (See  Exercise  26.) 


Fig.  295. 


§  276.   PYRAMIDS  AND   CONES. 

81.  In  a  regular  hexagonal  pyramid  a  side  of  the  base  =  im,  and  the  height 
of  the  pyramid^  i.8m;   find  (z.)  the  total  surface;    (z'z.)  the  volume. 

82.  The  roof  of  a  summer-house  has  the  shape  of  a  regular  octagonal 
pyramid ;  its  height  =  15**,  and  a  side  of  its  base  =  9ft.     Find  the  cost  of  the 
boards,  iin  thick,  necessary  to  make  it,  at  2  cents  a  board  foot. 

83.  A  vat  has  the  shape  of  an  inverted  frustum  of  a  square  pyramid;    a 
side  of  the  upper  base  =  4™,  a  side  of  the  lower  =  2m.  and  the  depth  of  the 
vat  =  5™.     How  much  water  will  it  hold? 

84.  The  pyramid  at  Ghiza,  near  Memphis  in  Egypt,  is  about  144™  high; 
the  base  is  a  square  whose  side  =  187™,  and  the  top  also  a  square  whose  side 
=  3.7™.     If  the  pyramid  were  solid,  what  would  be  its  volume? 

85.  Divide  a  pyramid  24°'"  high  into  two  equal  parts  by  a  plane  parallel 
to  the  base. 

86.  Find  (z.)  the  total  surface,  (zY.)  the  volume,  of  a  regular  tetrahedron 
whose  edge  =  8ocm. 


308  GEOMETRY    FOR    BEGINNERS.  [§   276. 

87.  The  base  of  a  pyramid  is  a  square,  and  its  sides  are  equilateral  tri- 
angles ;  find  its  volume  if  a  side  of  the  square  =  4Ocm. 

88.  The  edge  of  a  regular  octahedron  —  2m;  find  (z.)  its  surface;  (zz.)  its 
volume;  (zYz.)  the  ratio  of  its  volume  to  that  of  a  cube  having  the  same  edge. 

89.  A  hill,  having  the  form  of  a  cone,  is  to  be  removed,  and  the.  base 
divided  into  building-lots.     The  slant  height  of  the  hill=  120.8™,  and  the  cir- 
cumference of  Jhe  base  =  750™.     (z.)  How  many  cubic  meters  of  earth  must 
be  removed?     (zz.)  If  the  base  is  divided  into  six  equal  lots,  how  large  will 
each  be? 

90.  The  height  of  a  cone  =  86cm,  and  the  slant  height  is  twice  as  much; 
find  (z.)  the  area  of  the  base;  (z'z.)  the  volume  of  the  cone. 

91.  A  tower  has  a  conical  roof  5.8m  high  and  12.6™  in  diameter;  find  its 
area. 

92.  Three  leaden  cones,  each  4Ocm  high,  and  having  for  diameters  I2cm, 
24cm,  and  36°™,  respectively,  are  melted  together  and  cast  into  one  having  the 
same  height;  find  its  diameter. 

93.  If  the  three  cones  of  the  last  exercise  had  been  cast  in  the  shape  of  a 
cube,  what  would  be  the  edge  of  the  cube. 

94.  How  far  from  the  base  must  a  cone  of  sugar,  54cm  high  and  i8cm  in 
diameter,  be  cut  in  two,  parallel  to  the  base,  in   order  that  the  parts  may  be 
equal? 

95.  Show  how  the  cone  of  the  last  exercise  may  be  divided  by  planes  par- 
allel to  the  base  into  three  equal  parts. 

96.  Out  of  a  circular  piece  of  sheet-tin  86cm  in  diameter  a  sector  with  the 
angle  150°  is  cut,  and  this  is  then  rolled  into  the  shape  of  a  cone;  find  the 
volume  of  this  cone. 

97.  Explain  how  you  would  find  the  height  of  a  cone  having  given  the 
circumference  of  the  base  and  the  slant  height  ? 

98.  I  wish  to  find  the  value  of  a  straight  pine  tree  that  tapers  to  a  point. 
the  circumference  of  the  base  being  4m.     The  tree  casts  a  shadow  30™  Ion  ; 
at   the   same    time    that    the    shadow   of    a  vertical   rod  i.2m  long  is   i.G" 
What  is  the  tree  worth  at  $8  per  cubic  meter  (branches,  etc.,  being  regarded 
as  worth  the  price  of  felling  and  trimming)  ? 

99.  A  granite  column  has  the  shape  of  the  frustum  of  a  cone;   the  base  is. 
2.i2m  in  circumference,  the  top  i.5m,  and  the  height  =  5m.     Find  its  weight. 

100.  A  vessel  holding  I21  has  the  form  of  the  frustum  of  a  cone;   the 
lower  base  is  28cm  in  diameter,  and  the  upper  base  24°'";   find  its  height. 

101.  How  much  sheet  tin  is  required  to  make  a  speaking-tube  i.42mlong, 
and  43cm  and  36cm  in  diameter  at  the  ends? 


§277-]     CHAPTER  XIII. — SURFACES  AND  VOLUMES  OF  BODIES.  '        309 

102.  How  much  metal  is  required  to  make  a  tin  can,  the  diameter  of  the 
bottom  being  36cm,  that  of  the  open  top  i6cm,  and  the  height  being  48cm? 

103.  Assuming  the   cask  {Fig.  296}  to  consist  of  tw^o  frustums  of  cones 
placed  with  their  greatest   bases  together  at  the 

middle,  find  its  volume,  having  given  that  the  di- 
ameter of  the  middle  section  =  76°™,  the  diameter 
of  each  end  =  66cm,  and  length  =  im.      Will  the      j 
answer  obtained  be  too  large  or  too  small? 

104.  A  more  accurate  value  for  the  volume  of 

a  cask  may  be  found  by  use  of  the  formula,  pjg%  2g6. 

Volume  =  i  TT  h  (r1  +  2  ^), 

where  h  —  the  length,   r  —  radius  of  one  end,  R  —  radius  of  the  greatest 
section.     Find  the  volume  of  the  preceding  cask  by  means  of  this  formula. 

105.  How  many  bottles   of  wine,  each  holding  0.6^,  can  be   filled  from 
a  cask  in  which  the  greatest  and  the  least  diameters  are  86cm  and  62cm,  respect- 
ively, and  the  length  i.5m? 

106.  How  many  gallons  will  a  cask  hold  whose  length  is  3^,  and  greatest 
and  least  circumferences  I2ft  and  9ft,  respectively? 


§  277.    THE  SPHERE. 

107.  How  many  bullets  15™™  in  diameter  can  be  cast  from  ioks  of  lead? 

108.  The  English  national  debt  is  about  $4,500,000,000.     What  would  be 
the  diameter  of  a  sphere  of  gold  having  the  same  value,  assuming  iks  of  gold 
as  worth  $720? 

109.  The  diameter  of  the  sun  is  about  113  times  that  of  the  earth.     Find 
the  volume  of  the  sun,  assuming  the  diameter  of  the  earth  to  be  8000  miles. 

110.  Eighty  bullets,  equal  in  size,  are  thrown  into  a  cylindrical  vessel 
68cm  in  diameter  containing  water.     If  the  level  of  the  water  rises  2cm,  find 
the  diameter  of  a  bullet. 

111.  A  body  sinks  in  water  till  the  weight  of  the  water  displaced  —  weight 
of  the  body.     How  heavy  is  a  sphere  o.2m  in  diameter  which  floats  half  under 
water? 

112.  Find  the  weight  of  a  wooden  sphere  0.36™  in  diameter  if  one-sixth 
of  its  bulk  is  below  the  water-level  when  it  floats  on  water. 

113.  If   a  wooden    sphere   2Ocm  in  diameter  is  reduced  to  one-half  its 
original  volume  in  a  lathe,  what  will  then  be  its  diameter? 


310  GEOMETRY    FOR    BEGINNERS.  [§   278. 

114.  Find    the   diameter  of   a  sphere   made   by  melting   together   three 
spheres  whose  diameters  are  i.2m,  o.8m,  and  o.4m  respectively. 

115.  A  sphere  o.£m  in   diameter  is  reduced   in  size  by  turning  till  its 
diameter  is  one-twelfth  less  than  at  first.     How  much  is  the  volume  reduced? 

116.  The  outer  circumference  of  a  hollow  sphere  =  i.4m,  the  thickness 
=r  24mm.     Find  the  interior  volume. 

117.  An  arch  in  the   form   of  a  hemisphere   has  an  inner  diameter  of 
4.8™,  and   its  thickness  —  0.7™.     Find  how  many  cubic  meters   of   stone  it 
contains. 

118.  What  is  the  value  of  a  hemispherical  copper  kettle  im  inside  diame- 
ter, if   the    thickness    of   the   copper  =  icm,  and  copper  is  worth  $0.75  per 
kilogram  ? 

119.  The  diameter  of  a  balloon  =  30™;   the  material  of  which  it  is  made 
weighs  o.5ks  per  square  meter;   and  the  car  attached  below  weighs  40^.     Find 
its  ascensional  force  if  air  weighs  1.38  per  liter,  and  the  balloon  is  filled  with 
coal  gas  which  is  half  as  heavy  as  air. 

Hint.  —  The  ascensional  force  =  weight  of  the   air  displaced  by  the  balloon 
—  weight  of  the  balloon,  the  car,  and  the  gas  in  the  balloon. 

120.  What  must  be  the  outer  diameter  of  a  hollow  sphere  25mm  thick,  in 
order  that  it  may  hold  just  60  liters? 

121.  How  many  bodies  as  large  as  the  moon  could  be  made  from  the 
earth,  taking  the  diameter  of  the  latter  body  as  four  times  that  of  the  former? 

122.  If  a  drop  of  soap  and  water  4cm  in  diameter  is  blown  into  a  bubble 
I5cm  in  diameter,  find  the  thickness  of  the  bubble. 

123.  The  two  frigid  zones  contain  0.082  of  the  earth's  surface.     What  is 
the  height  of  each  zone?     What  is  the  diameter  .of  the  Arctic  circle? 


§  278.     EQUIVALENT   SURFACES   AND   SOLIDS. 

124.  A  cube  and  a  sphere  have  the  same  volume;  which  has  the  greater 
surface? 

125.  A  cube  and  a  sphere  have  the  same  surface;  which  has  the  greater 
volume  ? 

The  radius  of  the  base  of  a  cone  =  iocm,  the  height  of  the  cone  —  2Ocm. 
Find,— 

126.  The  edge  of  an  equivalent  cube. 

127.  The  height  of  an  equivalent  cylinder,  etc.,  having  the  same  base. 

128.  The  radius  of  an  equivalent  sphere. 


§   278.]     CHAPTER  XIII. — SURFACES  AND  VOLUMES  OF  feODIES.  31 1 

129.  How  high  must  a  right  prism  be,  whose  base  is  a  rectangle  i6cra 
X  I2cm,  that  it  may  be  equivalent  to  a  cube  whose  edge  =  iocm? 

130.  A  cube  whose  edge  =  iocm  is  cut  into   two  equal  parts  by  a  plane 
parallel  to  the  base,  and  the  parts  are  then  placed  with  the  half  sides  together 
so  as  to  form  a  prism.     Compare  this  prism  and  the  cube  as  regards  (z.)  vol- 
umes ;    (zY.)  surface  ;    {in.}  bases  ;    (z'?y.)  heights. 

Note  i.  —  Equivalent  prisms  have  not  always  equivalent  surfaces. 

Note  2. —  In  equivalent  prisms  (also  cylinders,  pyramids,  and  cones),  the  bases 
are  inversely  proportional  to  the  heights  (that  is,  the  less  the  base  fas  greater  the 
height,  and  conversely,  so  that  the  product  of  the  two  factors,  base  and  height, 
shall  be  always  equal). 

131.  In  two  equivalent  cylinders  the  ratio  of  the  bases  is  3:  5.     What  is 
the  ratio  of  the  heights  ? 

132.  The  dimensions  of  a  cylinder  are :  diameter  of  the  base  =  7cm,  height 
_  i2cm.     Find  the  height  of  an  equivalent  cylinder,  the  diameter  of  its  base 
being  6cm. 

133.  Compare  the  surfaces  of  the  two  bodies  in  Exercise  129.     Which  is 
the  greater,  and  by  how  much  ? 

Note.  —  Among  all  equivalent  four-sided  right  prisms  the  cube  has  the  least 
surface. 

134.  A  cube  with  an  edge  of  8cm  has  the  same  surface    as  a  right  prism 
whose  base  is  a  rectangle  6cm  X  4cm;   compare  their  volumes.     Which  is  the 
greater,  and  by  how  much? 

Note.  —  Among  all  four-sided  right  prisms  with  equivalent  surfaces  the  cube 
has  the  greatest  volume. 

135.  The  base  of  a  square  pyramid  is  6cm  X  6cm,  and  the  height  =  I2cm; 
find  the  height  of  an  equivalent  prism  with  an  equal  base. 

136.  Transform  a  cylinder  iocm  in  diameter  and  I2cm  high  into  an  equiva- 
lent square  pyramid  with  a  base  8cm  X  8cm. 

137.  Transform  a  cube  with  an  edge  of  8cm  into  an  equivalent  cylinder 
the  diameter  of  whose  base  shall  be   iocm;   then  compare  the  surfaces  of  the 
two  bodies. 

138.  Transform  the  above  cube  into  a  cylinder  with  an  equivalent  surface, 
the  diameter  of  its  base  to  be   iocm;   then  compare  the  volumes  of  the  two 
bodies. 

139.  Transform  a  cylinder,  in  which  the  diameter  of  the  base  and  the 
height  are  each  equal  to  iocm,  into  an  equivalent  cylinder  having  a  height  of 
i6cm;   then  compare  the  surfaces  of  the  two  cylinders. 

Note.  —  Among  equivalent  cylinders  that  has  the  least  surface  in  which  the  diam- 
eter of  the  base  and  the  height  are  equal. 


312  GEOMETRY    FOR   BEGINNERS.  [§   279. 

140.  Slant  height  of  a  cone  =  8cm,  and  diameter  of  its  base  =  6cm;  find  the 
height  of  a  cylinder  4cm  in  diameter,  having  an  equivalent  convex  surface. 

141.  Transform  a  sphere  6cm  in  diameter  into  an  equivalent  cylinder,  the 
diameter  of  its  base  to  be  the  same  as  that  of  the  sphere  ;   then  compare  their 
surfaces. 

142.  A  leaden  sphere  9cm  in  diameter  is  recast  in  the  shape  of  a  cube; 
find  the  edge  of  the  cube. 

143.  Another  sphere  of  the  same  size  is  recast  in  the  form  of  a  cylinder 
5cm  in  diameter;   find  the  height  of  the  cylinder. 

144.  What  must  be  the  height  of  a  cylinder  whose  base  is  4cm  in  diame- 
ter if  it  has  the  same  surface  as  that  of  a  sphere  4em  in  diameter? 

145.  A  sphere  is  iocm  in  diameter,  and  the  base  of  an  equivalent  cylinder 
has  the  same  diameter  ;   compare  the  surfaces  of  the  two  bodies. 

Note.  —  Among  all  equivalent  bodies  the  sphere  has  the  least  surface. 

146.  A  sphere  and  a   cylinder  have   equivalent   surfaces;    the  sphere  is 
locm  in  diameter,  and  so  likewise  is  the  base  of  the  cylinder ;   compare  the 
volumes  of  the  two  bodies. 

Note.  —  Among  all  bodies  having  equivalent  surfaces  the  sphere  has  the  greatest 
volume. 

§  279.   RATIOS  OF  SURFACES  AND  OF  SOLIDS. 

147.  Two  prisms  have  bases  each  containing  36(im;   what  is  the  ratio  of 
their  volumes  if  their  heights  are  iocm  and  I5cm? 

148.  Compare  in  general  the  volumes  of  two  prisms  if  they  have  equiva- 
lent bases. 

149.  Compare  in  like  manner  the  volumes  of  two  cylinders,  two  pyramids, 
and  two  cones,  with  equivalent  bases. 

150.  The  volumes  of  two  prisms  with  equivalent  bases  are  36ocbm  and 
56ocbra;  what  is  the  ratio  of  their  heights?     If  the  height  of  one  is  ay0'11,  find 
that  of  the  other. 

151.  A  cone  8cm  high  is  reduced  in  size  till  the  height  is  3.5cm,  the  base 
remaining  the  same.     What  fractional  part  of  the  whole  is  left? 

152.  Into  the  ends  of  a  cylinder  I2cm  long  two  cone-shaped  boles  3"" 
deep  are  bored.     If  the  cylinder  weighed  at  first  6ooS,  what  will  it  weigh  after 
the  holes  are  made? 

153.  If  a  cylinder  and  a  cone  have  equal  bases,  what  must  be  the  ratio  of 
their  heights  in  order  that  their  volumes  may  be  equal? 

154.  The  heights  of  two  prisms  are  each  2Ocm;   compare  their  volumes  if 
the  base  of  one  is  four  times  that  of  the  other. 


§   279-]     CHAPTER  XIII. — SURFACES  AND  VOLUMES  OF  BODIES.  313 

155.  Compare  in  general  the  volumes  of  two  prisms,  two  cylinders,  two 
pyramids,  or  two  cones,  having  equal  heights. 

156.  A  cone  6cm  iu  diameter  is  turned  in  a  lathe  until  the  diameter  is  re- 
duced to  5cm,  the  height  remaining  unchanged;   what  part  of  the  whole  cone 
is  left? 

157.  If  the  above  is  turned  till  it  is  reduced  to  one-fourth  its  original  vol- 
ume, what  will  be  its  diameter? 

The  edge  of  a  cube  =  iocm;   find,  — 

158.  Volume  of  the  largest  cylinder  that  can  be  made  from  it. 

159.  Volume  of  the  largest  sphere. 

160.  Volume  of  the  largest  cone. 

161.  Compare  the  volumes  of  the  above  cube,  cylinder,  sphere,  and  cone. 
The  diameter  of  a  cylinder  =  its  height  =  iocm;   find,  — 

162.  Volume  of  the  largest  sphere  that  can  be  turned  from  it; 

163.  Volume  of  the  largest  cone. 

164.  Compare  the  volumes  of  the  above  cylinder,  sphere,  and  cone. 

165.  What  part  of  a  hemisphere  is  the  largest  cone  that  can  be  made 
from  it? 

166.  If  the  diameter  and  the  height  of  a  cylinder  are  equal,  compare  the 
convex  surface  with  the  areas  of  the  bases. 

167.  If  the  diameter  and  the  height  of  a  cone  are  equal,  compare  the 
convex  surface  with  the  area  of  the  base. 

168.  If  the  convex  surface  of  a  cylinder  —  sum  of  areas  of  the  bases,  com- 
pare the  height  with  the  radius  of  the  base. 

169.  I  want  a  cylinder  whose  convex  surface  shall  be  six  times  the  area  of 
the  base.     Find  the  ratio  of  the  height  to  the  radius  of  the  base. 

170.  What  relation  must  exist  between  the  height  of  a  cone  and  its  diam- 
eter in  order  that  the  convex  surface  may  be  equal  in  area  to  the  base? 

171.  Compare  the  convex  surfaces  of  a  cylinder  and  a  cone  having  equal 
bases  and  heights.     Compare,  also,  their  total  surfaces. 

172.  Compare  the  surfaces  of  two  spheres  whose  diameters  are  6cm  and 
7cm.     What,  in  general,  is  the  relation  between  the  surfaces  of  spheres  differ- 
ing in  diameter? 

173.  A  cylinder  and  a  hemisphere  have  equal  bases  and  heights.     Com- 
pare (z.)  their  convex  surfaces  ;  (z'z.)  their  total  surfaces  ;   (m.)  their  volumes. 

174:.  The  diameter  of  a  cylinder  =  its  height  =  iocm;  find  the  volume  of 
the  largest  rectangular  parallelopiped  that  can  be  made  from  it. 

175.  Compare  the  volumes  of  a  sphere  and  the  largest  cube  that  can  be 
made  from  it. 


3 14  GEOMETRY    FOR    BEGINNERS. 


REVIEW  OF   CHAPTER   XIII. 

I.  —  Surfaces  of  Bodies. 

§  253.  Surface  of  a  prism. 

§  254.  Surface  of  a  cylinder. 

§  255.  Surface  of  a  pyramid. 

§  256.  Surface  of  a  cone, 

§  257.  Surface  of  the  frustum  of  a  cone. 

§  258.  Surface  of  a  sphere. 

II.  —  Volumes  of  Bodies. 

§  259.  Definition  of  Volume.     Units  of  volume. 

§  260.  Measurement   of  volumes.     Analysis    of  the    different   cases    to    be 

considered. 

§  261.  Volume  of  the  rectangular  parallelepiped. 

§  262.  Volume  of  the  right  parallelepiped. 

§  263.  Volume  of  the  oblique  parallopiped. 

§  264.  Volume  of  the  triangular  prism. 

§  265.  Volume  of  any  prism. 

§  266.  Volume  of  the  cylinder. 

§  267.  Volume  of  the  triangular  pyramid. 

§  268.  Volume  of  any  pyramid. 

§  269.  Volume  of  the  cone. 

§  270.  Results  of  §§  261-269  summed  up  in  two  rules. 

§  271.  Volume  of  a  sphere. 

§  272.  Relation  between  volume  and  weight. 

§  273.  Equivalent  solids,  similar  solids,  equal  solids. 

§  274.  Two  laws  applicable  to  similar  solids. 

III.  —  Exercises  and  Applications. 

§  275.  Prisms  and  cylinders. 

§  276.  Pyramids  and  cones. 

§  277.  The  sphere. 

§  278.  Equivalent  surfaces  and  solids. 

§  279.  Ratios  of  surfaces  and  solids. 


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